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OUTLINES OF PHYSICS 


A CONCISE DISCUSSION OF THE LAWS AND RELATIONSHIPS OF 
PHYSICS AND THEIR MATHEMATICAL EXPRESSION. 


BY 

CHARLES C. BIDWELL, Ph D. 

Vi 

Assistant Professor of Physics 
Cornell University 


Ithaca, N. Y. 

THE COMSTOCK PUBLISHING CO. 
1922 



QC 2.1 

,35 


COPYRIGHT 1922 

BY THE COMSTOCK PUBLISHING COMPANY 
ITHACA, N. Y. 


Press of W. F. Humphrey 
Geneva, N. Y. 

NOV 27 ’22 

© Cl A 6 92068 


PREFACE 


This text is a development from notes used for several years in 
mimeograph form by students of engineering and chemistry at 
Cornell University as a basis of a second course in physics. It 
is designed for those students who h^ve completed an intro¬ 
ductory course in college physics, and have had a first course in 
differential and integral calculus. A great deal of the descriptive 
material properly found in introductory texts is omitted. It is the 
purpose of the text to study the fundamental laws of physics, and to 
show how these laws are expressed in mathematical form. It 
is hoped thereby to bridge the gap between the introductory 
descriptive texts, and the more advanced mathematical treatises 
on special topics. 

It is regretted that through lack of time certain material on 
sound, which is in course of preparation, has had to be deferred 
for a future edition. 

October, 1921. C. C. B. 






CONTENTS 

MECHANICS 


PAGE 

Fundamental Concepts of Position and Motion. 

Displacement and Vector Quantities. Vector Operations. 

Velocity and Acceleration. Uniformly Accelerated Motion. 

Thi Motion of a Projectile. Uniform Circular Motion. 

Problems. 1 

Newton’s Laws of Motion. 

The First Law. The Second Law. Momentum. Impulse 

of a Force. The Third Law. 6 

Systems of Units. 

The c. g. s. System. The English Gravitational System. 

Dimensions. Problems. 8 

Simple Harmonic Motion. 

Definition. Solution of the Equation of S. H. M. Superposition 
of Simple Harmonic Motions. Expressions for Acceleration, 

Period, Force. Problems. 12 

Work and Energy. 

Mathematical Expression for Work. Units. Power. Graphi¬ 
cal Representation of Work. The Stretched Spring. Energy. 

Problems. 17 

Angular Motion. 

General Equations. Vector Representation of Angular Quanta 

ties. Moment of Force. Problems. 23 

Center of Mass. 

Definition. Velocity and Acceleration of Center of Mass. 

Action of a Couple. Motion of a Rigid Body under Action 
of a Single Force not through the center of Mass. Center of 

Gravity. Problems. 25 

Moment of Inertia. 

Definition. Angular Momentum. Impulse of a Torque. 

Work and Energy in the Case of Angular Motion. Moments of 
Inertia about Parallel Axes. Motion of a Rolling Body. 

Equation of Motion in the Case of Rigid Bodies. Problems... 29 

Gyroscopic Motion. 35 

Elasticity. 

Definitions. Types of Strain. Limitations of Hooke’s Law. 

The Twisted Shaft or Wire. Problems. 37 

HEAT 

Thermometry. 

Temperature Scales. Expansion. Problems. 41 











The Fundamental Gas Laws and the Equation of State. 

Boyle's Law. Charles’ Law. The Equation of State. The 

Absolute Zero. Work on the (p. v) diagram. Problems. 

The Kinetic Theory of Gases. 

Assumptions of the Kinetic Theory. Pressure and Temperature 
in terms of the Kinetic Theory. Avagadro’s Law. Brownian 
Movements. Deviations from Boyle’s Law. Van der Waal’s 
Equation. The Porous Plug Experiment. Mechanical Refrig¬ 
eration . 

Heat as Energy. 

Quantity of Heat. The First Law of Thermodynamics. 

Problems......*.. t . 

Change of State. 

Vaporization. Fusion. Problems. 

Properties of Gases. 

Application of the First Law to Gases. The Specific Heats of 
a Gas. Isothermal Changes. Adiabatic Changes. Problems. 
Thermodynamics. 

Cyclic Processes; Reversible and Irreversible Processes. 
The Carnot Cycle. Efficiency of a Carnot Engine. The 
Second Law of Thermodynamics. Carnot’s Theorem. The 
Thermodynamic Temperature Scale. Entropy. The Entropy- 
Temperature Diagram. Comparison of the Perfect Gas Scale 
with the Thermodynamic Scale. Deviations of Gas Thermom¬ 
eters from the Thermodynamic Scale. The use of Thermo¬ 
dynamic Equations. Maximum Efficiency of a Heat Engine. 

Problems. 

The Transfer of Heat 

The Radiation Laws and the Measurement of High Tem¬ 
peratures. Newton’s Law of Cooling. Kirchhoff’s Law. 
Uniform Temperature. Enclosures. Stefan’s Law. Wien’s 
Displacement Law. Wien’s Energy Distribution Law. Opti¬ 
cal Pyrometry. 

LIGHT 

The Nature of Light. 

Light as Wave Motion. Huyghen's Principle; Rectilinear 

Propagation. Color. 

Reflection. 

Reflection at a Plane Sin-face. Image in Plane Mirror. Re¬ 
flection at Spherical Surfaces. The Sagitta Theorem. Con¬ 
jugate Focal Relations for Concave Spherical Mirror. Convex 

Mirror. The Parabolic Mirror. Spherical Aberration. 

Refraction. 

Snell’s Law. Critical Angle. Refraction through a Plate with 
plane parallel Faces. Refraction through a Prism. The 
Spectrometer... 










PAGE 


Dispersion. 

Dispersion and Dispersive Power. Dispersion without Devia¬ 
tion; Direct Vision Spectroscope. Deviation without Disper¬ 
sion. Problems. 90 

Spectra. 

Cont inuous Spectra. Wave-Length Units; The Spectral Range. 

Bright Line Spectra. Absorption Spectra. The Color of 

Bodies. Luminescence. 91 

Lenses. 

Refraction through Lenses; Types of Lenses. Path of Ray 
through a Lens; Construction of Image. Conjugate Focal 
Relations; Huyghen’s Construction for Converging Lens. 

Rule of Signs. The Diverging Lens. Spherical and Chromatic 


Aberrations. Combinations of Lenses. Achromatic Com¬ 
bination . 94 

Optical Instruments. 

Lens as Magnifier. Simple Compound Microscope. Astron¬ 
omical Telescope. Galileo’s Telescope. 99 

Interference; Diffraction. 

The Zone Plate. Diffraction passed a Straight Edge. The 

Diffraction Grating. 102 

Polarization. 

Double Refraction and Polarization. Polarization by Re¬ 
flection. The Nicol Prism; the Polariscope. Rotation ofthe 
Plane of Polarization. 107 


ELECTRICITY 

Electrostatics. 

Conlomb’s Law. Electric Field and Electric Induction. 

Gausses Theorem. Potential. Capacity. Condensers. 

Dielectric Constant. Units. Capacity of a Sphere. Capaci¬ 
ty of a Plate Condenser. Capacity of a Spherical Condenser. 

Capacity of a Cylindrical Condenser. Capacity of Condensers 
in Parallel and in Series. Energy of a Charged Condenser. 112 

Magnetism. 

Conlomb's Law. Magnetic Field and Magnetic Induction. 

Gausses Theorem. Induction and Field in a Transverse 
Gap in an Iron Ring. Induction and Field in a Transverse 
Gap in a Straight Iron Bar. Case of Bunched Windings on an 
Iron Ring. Case of Field due to Current in a Wire. Perme¬ 
ability. Torque experienced by a magnet. Magnetic Poten¬ 
tial. Problems. 120 

The Electric Current and its Effects. 

Production and Detection of Electric Current. Side Push on 
a Wire carrying Current. Magnetic Field due to Electric 









PAGE 


Current. Magnetic Field at the Center of a Circular Loop. 
Units. The Field on the Axis of a Circular Coil. The Field 
near a Long Straight Wire. Field at Center of a Solenoid. 
Work to Carry Unit Pole around a Wire Carrying Current. 


Quantity of Electricity. 127 

Measurement of Current. 

The Tangent Galvanometer. Torque on a Coil Suspended 

in a Magnetic Field. The D’Arsonval Galvanometer. 132 


Ohms's Law. 

Joules Law; Resistance. Ohm's Law for a Portion of a 
Circuit. Ohm's Law for a Complete Circuit. General Expression 
for Ohm's Law. Kirchhoff’s Laws. Application of Ohm's 

Law to a Simple Circuit. 134 

Measurement of Resistance and E. M. F. 

The Wheatstone Bridge. The Potentiometer. Voltmeters 
and Ammeters. Resistances in Parallel. Effect of Tem¬ 
perature on Resistance. Power and Energy in the Electric 


Circuit. 138 

Induced Electromotive Forces. 

Lenz’s Law. Alternating Current Dynamo. Direct Current 
Dynamo. The Direct Cm rent Motor. Efficiency of a Motor. 

The Induction Motor. 140 

Self and Mutual Induction. 

Self Induction. Mutual Induction. Ohm’s Law for a 

Circuit Containing Inductance. 147 

The Ballistic Galvanometer and Measurement of Capacity. 

The Ballistic Galvanometer. Logarithmic Decrement. 149 

Magnetic Induction in Iron. 


Effect of Poles. Hysteresis. Experimental Methods of 
Obtaining the Hysteresis Loop. Work to Magnetize a Sample 
of Iron. Relation between Flux of Induction and Ampere 
turns. The Magnetic Circuit. Method of Solving Problems 

on the Magnetic Circuit. 153 

Electrolysis. 

Conduction of Electricity. Faradays Laws of Electrolysis. 

Energy Equation of an Electrolytic Cell. The Voltaic Cell. 

Theory of the Chemical Action of a Simple Voltaic Cell. 

Energy equation of a Voltaic Cell. Source of Electrical 


Energy in a Daniell Bell. Secondary Cells. 160 

Thermo-Electricity. 165 

Alternating Currents. 


The Fundamental Equations of the Alternating Current 
Circuit. Circuit with Inductance and Resistance—The 
Equation for Current. Vector Representation of Voltages in an 
A. C. Circuit. Circuit with Inductance, Resistance and 










PAGE 


Capacity. Vector Diagram for Circuit with Capacity. 

Average current and E. M. F. Virtual Current and E. M. F. 

Power in an A. C. Circuit; Power Factor. The Wattmeter. 
Application of Ohm’s Law to A. C. Circuits. Addition 
of Electromotive Forces out of Phase. Addition of Currents. 
Impedances in Series. Impedances in Parallel. Inductance 
and Capacity in Series with Resistance negligible. Oscillating 

Circuits. Coupled Circuits; the Transformer. 166 

Electromagnetic Radiation. 

Electric Displacement. Displacement Current. Electric 
Oscillations and Radiation. The Detection of Electromag¬ 
netic waves. The Three-Electrode Vacuum Tube. The 
Vacuum Tube as a Detector. The Vacuum Tube as an 
Amplifier. The Vacuum Tube as a Generator. Heterodyne 
Reception by Vacuum Tubes. Radio-Telephony; Voice 
Modulation. 182 







MECHANICS 


FUNDAMENTAL CONCEPTS OF POSITION AND 
MOTION 


1. Displacement and Vector Quantities. The position of a 
particle with reference to some arbitrary origin is usually desig¬ 
nated either by means of rectangular coordinates x, y and z, or 
by means of vectors. The line drawn from the origin to the 
particle is called the position vector , s. When the position 
of a particle is changed, the displacement is given by specifying 
the changes Ax, Ay, Az or by As, the change in the position 
vector. All physical quantities may be classified as either scalar 
quantities or vectors , the former having magnitude only, the 
latter magnitude and direction. A vector quantity may be 
represented by a line whose length and direction indicate re¬ 
spectively the magnitude and direction of the quantity. Thus 
in addition to the position vector, we have the displacement 
vector which is the line drawn from the point representing the 
initial position of a particle to the point representing the final 
position; we have velocity vectors, lines specifying direction 
and magnitude of velocities; acceleration vectors, force vectors 
and various others. 

2. Vector Operations. If a particle is subjected to two or 
more displacements the resultant displacement may be determined 

graphically by laying off the 
corresponding vectors end to 
end and drawing the line 
from the initial to the final 
position as the resultant 
displacement. Thus the 
sum of displacements oa and 
ab (Fig. 1) is the displace¬ 
ment ob. The resultant 
displacement may also be 
computed by trigonometry 
since the vectors form the 
two sides of a triangle with 
lengths and directions 
known. The above method 
of addition applies in general to any like vectors. If several 
like vectors when placed end to end form a closed polygon the 
resultant vector, that is, the sum of all the vectors, is zero. (Fig. 2). 
This is a test which may be applied in the case of balanced forces; 
the sum of all the forces vectorially added must give a closed 
figure. 





2 


MECHANICS 


A single vector may be resolved into any number of component 

vectors the only condition 
being that the sum of the 
components vectorially added 
equal the original vector. The 
vector ob (Fig. 1) may be 
resolved into vectors oa and ab 
and is equivalent to these two. 
We are often concerned in 
physics with the multiplication 
and division of vectors. The 
product or quotient of parallel 
vectors is a scalar. Thus the 
product of force and distance 
traveled in the direction of the 
force is the scalar quantity 
work. Force divided by area 

gives the scalar quantity, pressure. 

The product or quotient of two mutually perpendicular vectors 
is a third vector at right angles to the two. Thus the product of 
two sides of a rectangle is the vector quantity area, the direction 
of an area being given by the normal to it. The product of force 
and lever arm is moment of force or torque, indicated by a vector 
along the axis of rotation. An angular displacement measured in 
radians may be considered as a linear distance along an arc divided 
by a radius, thus giving radians and indicated by a vector along 
the axis of rotation. 

3. Velocity and Acceleration. If, s, represents the displace¬ 
ment of a particle from some origin (in such a case the displace¬ 
ment and position vectors are the same), then ds/dt, the rate of 
change of position with time at some particular instant is called 
the velocity, v, at that instant. Magnitude, direction and sense 
must be expressed in specifying a velocity. (By sense is meant 
the sign, away from the origin usually being positive). 

If, v, represents the velocity of a particle then dv /dt, the rate 
of change of velocity with time at a particular instant, is called 
the acceleration, a, at that instant. Magnitude, direction and 
sense must here also be specified. In mathematical language we 
may express these definitions as 

v = ds /dt 1 

a = dv/dt = d 2 s/dt 2 2 

Equation 1 may be written 

ds = vdt 

f rom which by integration 


« 

e 




FUNDAMENTAL CONCEPT 3 

s = f vdt 3 

Likewise for 2 

v = f adt 4 

Equations 1 «rrd 4 are general equations fitting any type of 
motion. 

4. Uniformly Accelerated Motion. Motion in which the 
acceleration is constant is said to be uniformly accelerated motion. 
In this case the integration of equation 4 gives 

v = at + c 

where c is the constant of integration. The physical meaning 
of c is found by introducing the boundary conditions. Thus it 
is evident that c = v 0 , i. e., the velocity when t = 0 or in other 
words, the initial velocity. Therefore we have 

v = at + v 0 

If this value of v be substituted in equation 3 we have 

s = a f tdt + v 0 f dt 6 

Upon integration this gives 

s = i at + v 0 t + c' 7 

c' is the constant of integration. Interpreting as before, we 
find c' = s 0 . Therefore * 

1 2 

s = - at + v 0 t + s 0 

QUESTIONS 

1. Following the method employed in the case of uniformly 
accelerated motion discuss the case where a = 0, (uniform motion) 
finding v and s. 

2. Draw curves representing (a) velocity and (b) distance 
covered from starting point as a function of time, (body initially 
at rest) for 

(a) acceleration = 0 

(b) acceleration = const. ( = a) 

(c) acceleration = const, x time (= ct) 

(d) acceleration = — A sin pt (where p and A are constants). 
(Suggestion: In each case find the expression first for velocity 

and then for displacement by integrating equations 4 and 3 


4 


MECHANICvS 


respectively. Assume convenient numerical values for the 
constants and plot curves with time as abscissa. In each of the 
above cases as in uniformly accelerated motion, the process of 
integration is merely that of anti-differentiation, that is, the 
integral of a given quantity is the function which when differ¬ 
entiated gives the quantity. There may be several specific 
functions which will satisfy this condition. It is necessary to 
find the most general function. For instance, the integral of 
equation 4 may be v = at, but this is the special case of starting 
from rest. The addition of the constant of integration makes the 
solution general. 

In substituting values in the equation for question (d) above, 
namely a = — A sin pt, and in the corresponding expressions for v 
and s if values of pt are found as multiples of t /6 the sin and 
cosine computations will be much simplified. Thus let pt = 0. 
7r/6, 27r/6, 3x/6, 47r/6, 57 t/ 6 etc. carrying this at least up 
to 24 tt/6). 

5. The Motion of a Projectile. Suppose a body is projected 
into the air with an initial velocity, v 0 , at an angle, 0, with the 

horizontal. Required to find 
expression for (a), the range , 
(the horizontal distance tra¬ 
veled when the body reaches 
the ground; (b), the time of 
flight; (c), the maximum 
height and (d) ,the equation of 
the path. It is convenient to 
resolve the motion into its horizontal and vertical components. 
The horizontal component is subjected to no acceleration (ig- 
noringair friction), the vertical component has an acceleration, 
—g, since upward is taken as positive. We may then find hori¬ 
zontal velocities, v x , and displacements ,s x , by applying the 
expressions for uniform motion; and vertical velocities, v y , 
and displacements, s y , by applying the expressions for uniformly 
accelerated motion. Thus, taking the origin at the starting 
point, we have for the horizontal component 

v x = v 0 cos 0 8 

$ > v 0 cos 0 * t 9 

For the vertical component 

v y = (-g)t + v 0 sin 0 10 

s y = -*(-g)t 2 + v 0 sin 0 • t. 



11 




FUNDAMENTAL CONCEPTS 


5 


The range, that is, the horizontal displacement when the pro¬ 
jectile has again reached the ground, may be found by substi¬ 
tution s y = 0 in equation 11, [why?], solving for t which 

gives the time of flight and substituting this value of t in equa¬ 
tion 9. At the instant of maximum height the vertical velocity 
is zero. The time for the body to attain its maximum height 
may therefore be found by substituting v y = 0 in equation 
10. This value substituted in equation 11 gives the value of 
maximum height. The equation of the path may be obtained by 
eliminating t between equations 9 and 11. The equation 
thus obtained may be shown to be a parabola. The student 
should obtain for himself the above expressions. 

PROBLEM 


In the case of a projectile show, (a), that the time of ascent and 
descent are the same; (b), that the initial and final vertical 
velocities are the same. 

6. Uniform Circular Motion. This is the motion of a particle 
traveling in a circle with constant speed. Suppose the particle, 

traveling around the circle 
with center at o, Fig. 4, is 
at a given instant, in position 

(1) and after the lapse of a 
small time, At, in position 

(2) , having traveled along 
the arc a distance, As. 
Let Vi and V 2 represent the 
two velocities equal in 
magnitude but differing in 
direction. The velocity 
vectors may be placed in 
juxta-position as shown, the 

vector Av being the change in velocity in the time At. Since 
the points (1) and (2) are very close together, the arc As may be 
considered as sensibly straight. The triangles r, r, As and Vi, 
v 2 , Av are similar (homologous sides being perpendicular). 
Hence 

Av/v = As/r 12 

Dividing each” side by At we have 

A v _ As 13 

v A t r A t 




Fig. 4 


A v _ dv 
At" dt 


a and =4 t = v, therefore 
At at 


In the limit 





6 


MECHANICS 


v 2 14 

r 

The vector diagrams indicate that the direction of the accelera¬ 
tion is perpendicular to that of the velocity at any instant, the 
direction of the velocity being along the tangent, and that of the 
acceleration being toward the center. 

PROBLEMS 

1. A train gains a speed of 40 miles per hour in 6 minutes. 
Find the average acceleration in miles per hour per second, and 
in miles per second, per second. 

2. A column with circular base is under construction. The 
height is increasing at the rate of 8 feet per day. How fast is the 
volume increasing? 

3. A weight is dropped from an aeroplane traveling northward 
at a speed of 40 miles per hour. The aeroplane is at an altitude 
of 1,000 feet. How long before the weight reaches the ground 
and what velocity does it acquire? What is the equation of the 
path and what sort of a curve does this represent? (Ignore air 

4. A gun which produces 
a projectile velocity of 200 
ft. per second is mounted 
aboard a car with its barrel 
at right angles to the di¬ 
rection of motion of the car, 
as shown in figure. The car 
is traveling 50 ft. per second. 
The sight would have to be 
arranged at an angle 6 to 
the gun barrel as shown. 
Find the value of 6 so that 
the ball may hit any object 
which, at the instant of 
firing is in the line of sights. 
(From Nichols and Frank- 
/ lin, Elements of Physics) 

1 5. A stone weighing 200 

Fi &- 5 . grams is whirled vertically in 

a sling, the cord of which is 100 cm. long. What is the minimum 
linear velocity that will prevent the stone from falling out? 

NEWTON’S LAWS OF MOTION 

7. The First Law. If a body is at rest it will continue at rest 
or if in motion it will continue in motion in the same straight line 


resistance m this problem.) 




/it 


dr 

rO 

/ 

/ 


c 

- ^ 







NEWTON’S LAWS OF MOTION 


7 


unless acted upon by some agency or influence external to the 
body which is said to cause the change. This agency, which 
must be acting when any change in the motion of a body occurs, 
is called a force. 

This law is based upon experimental evidence. It gives a 
qualitative definition of the term force, as the cause of any change 
in the motion of a body. 

•8. The Second Law. Experiment shows that for a given 
force the acceleration produced depends upon the mass upon 
which the force acts—the greater the mass the smaller the ac¬ 
celeration and vice versa. For example, consider a locomotive 
capable of exerting a definite constant pull.—With twice as long a 
train (twice the mass) it will take twice as long to attain a given 
speed (neglecting friction considerations). The acceleration 
thus varies inversely with the mass. For a given train two 
engines will develop the given speed in half the time—or the 
acceleration is twice as great—thus acceleration is proportional 
to force. 

For a given force experimental evidence thus shows that in¬ 
creasing the mass acted upon results in decreasing the acceler¬ 
ation in direct proportion and vice versa. 

In other words the product of mass times acceleration is constant 
and is a measure of the force acting. Hence the defining equa¬ 
tion 

F = k ma 15 

(k) is a proportionality constant. Its value depends upon the 
units. 

This law describes the behavior of a body when acted upon by 
an unbalanced force. It may be thus stated—“when acted 
upon by a force a body gains velocity in the direction of the 
force at a rate which is proportional to the force and inversely 
proportional to the mass.” Equation 15 is the fundamental 
equation of mechanics. 

9. Momentum. The product, mv, is called momentum. 
The rate of change of momentum of a particle, m, is a measure 
of the force acting upon it. That is, 

F = k ^(mv) 15 a 

This is simply another expression of equation 15. 

10. Impulse of a Force. The product of force times time is 
called the impulse of a force. In general the impulse of a force 
is J\Fdt. For a constant force impulse is Ft. The impulse of a 

force is proportional to the momentum acquired under the action 
of the force. This follows directly from equation 15. Multiply- 


8 


MECHANICS 


ing both sides by t, assuming the body originally at rest, 
Ft = kmat = kmv 16 

Or, if the force is a variable, multiplying equation 15 by dt and 
integrating, we have 

f Fdt = km f adt 
by definition adt = dv, hence 

Impulse = / Fdt = km /'dv = kmv 17 

If the body has an initial velocity v 0 the integration is between 
the values v and v 0 or 

Impulse = f Fdt = km f dv = k (mv-mvo) 18 

Or the impulse is proportional to the gain in momentum. 

11. The Third Law. A change of motion, or the effect of a 
force, can only be obtained through the mutual interaction of 
one body upon another. That is, the action of a force requires 
two bodies and the effect of a force is the same upon each but 
oppositely directed. In other words to every action on one 
body there is an equal and opposite reaction on some other body* 

PROBLEM 

Show from these laws that when two bodies collide the total 
momentum is unchanged. 

SYSTEMS OF UNITS 

12. The C. G. S. System. In this system three fundamental 
units are selected, namely, the centimeter, the gram and the 
second. We may have as many units as we have kinds of quanti¬ 
ties but these are called “ derived units ” since they may all be 
expressed in terms of the three fundamental units of length, 
mass and time. 

The gram mass is 1 /1000 of the mass of a certain block of 
platinum kept at the International Bureau of Weights and Measures 
at Sevres, near Paris, and known as the Kilogram prototype. 

The centimeter is 1 /100 of the distance between two marks on 
a platinum bar preserved also at Sevres. 

The second is 1 /86400 of the mean solar day. 

In this system the unit of force, called the dyne, is defined as 
that force which will accelerate a gram mass at the rate of one 


*Care should be taken to distinguish between (a) action and reaction and 
(b) cases of balanced forces. Thus, a book lies at rest on the table. The 
earth pulls on the book giving it its weight, the book also pulls with an equal 
force on the earth. This is action and reaction. Note that the action is on 
one body, the reaction on the other. The weight of the book is balanced by the 
upthrust of the table. Here are two balanced forces. They act on the same 
body. Each of these balanced forces has its action and reaction. 



vSYSTEMS OF UNITS 


9 


cm. per second, every second. We see that this is a derived unit 
since it is expressed in terms of the fundamental units of length 
and time. 

With this choice of units k in equation 15 becomes unity, 
and usually is not expressed. A system based upon the funda¬ 
mental units of length, mass and time in which an absolute force 
unit rather than a gravitational force unit is chosen is called an 
absolute system. 

Force is sometimes expressed in gravitational units, namely, 
in terms of the weight of the gram mass. We speak of a force of 
so many grams meaning so many times the weight of the gram 
mass. Whenever such a unit is used it should be noted that 
k in equation 15 is not unity but has the value, 1 /g. This 
follows from the fact that the weight of a gram mass is the pull 
of gravity upon it and that gravity accelerates a gram mass at 
the rate of g cm. per sec., per sec. Therefore, allowing the weight of 
one gram, as unit force, to act upon one gram mass we find ex¬ 
perimentally that a in equation 15 is g cm. per sec., per sec. and 
that therefore k = 1 /g. Such a system of units is called a gravi¬ 
tational system. 

Formulas in physics are usually derived on the basis of the 
absolute c.g.s. system in which the fundamental equation is 
F = ma. This fact should be remembered and substitutions in 
such formulas made accordingly, that is, forces should be con¬ 
verted to dynes. Throughout the work on mechanics in this 
text the k will not be dropped. Its presence will serve two 
purposes (1) to show the fundamental dependence of these formu¬ 
las on Newton’s Second Law and (2) to enable one to use any 
system of units he pleases, the value of k being selected to 
correspond. 

13. The English Gravitational System (Engineering System). 

In this system the weight at London of a certain mass of platinum 
(the pound mass) kept at the Exchequer in London is taken as 
the unit of force. The foot is the unit of length in physical 
measurements although legally the yard is the unit and is de» 
fined as the distance between two lines on a bronze bar kept at 
the Exchequer in London. In engineering computations the 
pound mass is usually not used as the unit, the term w/g being 
arbitrarily substituted for mass. 

If w/g is taken to mean mass we have unit mass when w/g = 
1. This says unit mass weighs (g) pounds. 

In this system the fundamental equation is written 


This is simply a statement of the proportionality between forces 
and accelerations. 


10 


MECHANICS 


In this system the mass of a man weighing 160 pounds is Jf, 1 
since mass = w/g (assuming g = 32). This unit of mass is 
sometimes called the “slug.” The engineering system is not an 
absolute one since mass does not enter as a fundamental unit. 

It should be noted that in the absolute c.g.s. system it is also 
true that w/g = m, (w) being of course expressed in absolute 
units (dynes). In fact, in any system in which k = 1, w/g = m. 

(An English absolute system, which has long been urged but 
not generally adopted in practice as yet, selects the three funda¬ 
mental units of length, mass and time, as in the c.g.s. absolute 
system, namely, the foot, the pound mass, and the second to¬ 
gether with an absolute unit of force, the latter being that force 
which will accelerate a pound mass at the rate of one foot /sec. /sec. 
This unit of force is called the poundal. k in this system is 
unity as it would be in any absolute system). 

14. Dimensions. In any absolute system all derived units 
may be expressed in terms of the three fundamental units of 
length, mass and time. The dimensions of a unit are the powers 
of the fundamental units to which it is proportional. For in¬ 
stance a velocity unit must express a length divided by time or 
its dimensions are (+1) in length, and (—1) in time. Acceleration 
is expressed in centimeters per sec. per sec. and its dimensions are 
(+1) i n length, (—2) in time. The dimensional formula for 
velocity is [L] [T] -1 , for acceleration [L] [T] ~ 2 . From equation 
15 we see that the dimensional formula for force is [M] [L] [T] ~ 2 , 
k being merely a number and having no dimensions. 

Dimensions are convenient in specifying a derived unit which 
has no name, thus both velocity and acceleration are expressed 
in terms of their dimensions, vel. as cms./sec., acceleration as 
cms. /sec 2 etc. Dimensions afford a valuable method of checking 
the accuracy of an equation. Both sides, to stand for the same 
thing, must be of the same dimensions. Thus suppose we have a 
case of potential energy converted to kinetic and we equate 
potential energy lost to kinetic energy acquired, obtaining such an 
equation as 

mgh=^ mv 

Expressed dimensionally this is [M] [Lf[T] " 2 = [M] [L] 2 [T] - 2 . 
The equation is therefore dimensionally correct. Every correct 
equation must satisfy such a relation. We sometimes are enabled 
to determine physical relations by means of this property of 
dimensional formulae. For example suppose we know from 
experiment that the velocity of sound in an elastic medium 
depends only upon two factors, the density (5) and elasticity of 
the medium. Density is mass per unit volume; the modulus of 


SYSTEMS OF UNITS 11 

elasticity (. M) concerned is the stretch modulus, which is force 
per unit area divided by elongation per unit length. The equation 
is 


Velocity = k 5 X M y 

where k is a proportionality constant and x and y are unknown 
powers of the density and elastic modulus respectively. 
Dimensionally this is 

[L] [T] -i ,= [ [M] [L] ‘ * * 3 4 5 6 ] ' ' [ [M] [L]-> [T] A 7 

From observation we see that the only values of x and y which 
will satisfy this equation are x = —y = +-. Hence 

Z — 

the preceding equation becomes 

v = k\/ M/8 


PROBLEMS 


1. A mass is clamped near the outer end of one of the spokes ot 
a wagon wheel. Describe the change in direction and magnitude 
of the force acting on the mass during one complete revolution 

(a) when the wagon is speeding up (assume uniform acceleration) 

(b) when it is going with uniform velocity, (c) when slowing down. 
Assume no friction 


2. M = 10 kg., m = 
100 g. Find acceleration of 

the “system”, (a) when 
0 = 90°, (b) when 0 = 

120°. Three seconds after 
starting the string is cut 
just above m. How far 
will M go in the next 5 
seconds ? 

3. A mass of 2 kgm is acted upon by a force of 5,000 dynes for 
3 minutes. No other force is acting. How far will the body go 
in that time, and what velocity will it acquire ? 

4. An elevator reaches full speed of 3 meters per sec. 2 sec. 
after starting. With what force in kg. wt. will a man “weighing” 
75 kg. push down on the floor, (a) when the elevator is starting 
upward, (b) when it is starting downward ? 

5. A cord is hung over a pulley. At one end is a mass of 10 
kg. At the other end a mass of 11 kg. Find the acceleration of 
each mass and the tension in the cord. Find the tension in the 
cord in problem 2. 

6. Two buckets, each weighing 500 lbs. are suspended from 

the ends of a rope passing over a windlass. A gallon (10 lbs.) of 




EJ 


Fig. 








12 


MECHANICS 


water is poured into one of the buckets. Find how far it will 
descend in 10 seconds, neglecting friction. 

7. a) A train has a mass of 350,000 kg. Starting from rest 
it reaches a speed of 2 yi kilometers per hour in 5 minutes. If 
the acceleration is uniform, what is the average pull of the 
locomotive in kg. wt.? (Assume that friction is constant and 
equals 4 kg. force per 1,000 kg. mass). 

b) The above train, moving at a speed of 50 km. per hr. is 
brought to a standstill in 16 seconds. Find the average pull of 
the locomotive in kg. wt. 

8. Solve problem 7 for a train moving up a ^ percent 
grade. 

9. A train starts from rest on a level line and moves through 
1,200 feet in the first minute. It then begins to ascend a uni¬ 
form incline, up which it is found to run with uniform velocity. 
Find the inclination of this portion of the line on the supposition 
that the engine exerts a constant pull. 

10. A locomotive is traveling on a curve. If the radius 

of the curve is r and the 
velocity (v) the mass 
(m), find the angle (6) 
of inclination such that 
no outward force is 

exerted against the rails by the flanges of the wheels. 

11. An 80 ton locomotive goes round a railway curve of which 
the radius is 500 ft. at a velocity of 50 ft. per sec. With what 
force in pounds weight do the flanges of the wheels of the loco¬ 
motive push against the outer rail, when the outer rail is not 
elevated? What is the proper elevation to give the outer rail, 
the width of track being 4 ft. 8 ^ inches. 




12. Find the velocity 
at which the belt must 
travel to exert no pressure 
on the pulleys. 


Fig. 8 


SIMPLE HARMONIC MOTION 

15. Definition. Simple Harmonic Motion is motion in which 
the acceleration is proportional to the displacement and in the 
opposite direction. Forces brought into play by elastic dis¬ 
tortions produce motion of this type, the return force and there¬ 
fore the acceleration being proportional to the distortion and 



SIMPLE HARMONIC MOTION 


13 


in the opposite direction. Examples of such motions are the 
vibrations of stretched springs, twisted wires, etc. 

The defining equation of S. H. M. for the case of a particle 
(linear motion) is, 

a = —ks or d 2 s/dt 2 = —ks 20 

Equation 20 is a differential equation involving two variables 
s and t. s can be expressed in terms of t and constants 
or s is said to be a function of t. The solution of the equation 
in this case is the particular function which, when differentiated 
twice, gives the original equation. 

16. Solution of the Equation of S. H. M. Eq. 20 may be 

written as 


d 2 s 

+ ks = 0 or a = —ks 


As general definitions we have 


ds 1 dv 
v ^and a 


From these we may write vdv = ads 
or for this case 

t- vdv = —ads 
k 

On integration this becomes 

1 v 2 s 2 
k2 = ~~2~ + C 

c is the constant of integration. We may write this 
v 2 = —ks 2 + kc'. 


c' must be the value of s 2 when v = 0. Let us call c' = s . 

o 


Then 


V = + -y/ k sj S 2 — S 2 

0 

ds 


If we substitute this in the equation v = we may write 


ds 


+ V S 2 -! 


\/k dt 


The integration of the two equations here expressed gives 

_i g _i s _ 

sin -= x/T t + e and cos ~ = Vk t -f e' 
o o 





14 

MECHANICS 


or 

s = s 0 sin ( VIF t + e) 

21 a 

and 

s = s 0 cos ( y/lTt + e') 

21 b 


17. Physical Meaning of these Solutions and Interpretation 
of the Constants. The solutions state that we can express the 
displacement as a sine or a cosine function of an angle which 
varies directly with the time. 

We may gain some light on this by considering the case of a 
particle on the rim of a wheel # rotating with constant angular 
velocity. Suppose co represents the angle swept over per 
second by a spoke (radius s 0 in Fig. 9). The angular dis¬ 
placement .(angle swept over) in time't, is cot. If a vertical line 
is dropped from the particle to the diameter, the displacement of 

the lower end of this line 
from the center of the 
circle, that is, the pro¬ 
jection of the motion of 
the particle on the diame¬ 
ter, is given by 

s = So sin (cot + e) 
or s = So cos (cot + e'). 
These equations are identi¬ 
cal in form with equations 
21a and 21b. The pro¬ 
jection of a uniform circu¬ 
lar motion is therefore 
simple harmonic. For 
a simple harmonic motion 
we may therefore con¬ 
struct a “circle of reference” indicating the path of a particle 
for which the S. H. M. may be considered the projection. We 
use the sine function if we measure the angle from the mid¬ 
position where s = 0, the cosine function if we choose to measure 
the angle from the end position where s = s 0 . e or e' repre¬ 
sents the angular displacement in the circle of reference at the 
time t = 0. The angle (cot + e) or (cot -f- e') is called the “phase 
angle” and gives the phase of the motion at time, t. e or e' 
is the phase angle when t = 0, i. e., when we start to count time. 
When cot attains the value 2tt , the motion has passed through 
all its phases and is about to start a new cycle. The time re¬ 
quired for a complete cycle is called the period , T, of the motion. 
Thus when t = 2tt, t = T and co = 27r/T. The maximum dis¬ 
placement (called the amplitude of the motion) is obviously 

So. 

Equations 21 a and 21 b may thus be interpreted in terms 







SIMPLE HARMONIC MOTION 


15 


of a “circle of reference.” 


Vk may be written as 


2tt 

T 


, for, at 


the time t = 0, s = Sp sin e, and at such time that Vk t = 2tt 
we have again s = So sin e, since sin (27r+e) = sine. That is, after 
such a time the displacement has again its initial value and the 
motion starts to repeat itself. This time is the period of the 

motion, T. Therefore \/k = We may now write the so¬ 
lution as 


s = s 0 sm ^ 


( 2ivt 


2irt \ 

T +V 


or s = s 0 cos 


/2irt 


f + O) 


22 a 


22 b 


In the latter case the reference position from which the phase 
angle is measured is 7r/2 radians behind the reference position 
for the first case. 

It should be understood that the circle of reference is merely 
an artifice of assistance in interpreting the equations. The 
phase angle has no physical significance as an angle in the circle 
of reference , for the circle of reference is purely imaginary. The 
equations simply state that the displacement varies with the 
time in the same way that a sine or cosine function of an angle 
changes when the angle varies directly with the time. The 
projection of a uniform circular motion on a diameter gives us a 
useful illustration of exactly this motion and enables us to visual¬ 
ize the phase angle. In strict terms the phase angle is merely 
the angle whose sine (or cosine) is s/s Q , i. e., the ratio of the dis¬ 
placement of any instant to the maximum displacement. 

18. Superposition of Simple Harmonic Motions. If we are 
considering a single motion we may select time, t, as zero, i. e., 
we may start counting time when the particle is passing through 
its mid-position. In this case the first equation simplifies to 

s = s 0 sin Or if we take t = 0, when the particle is in the ex¬ 


treme position, then, using the second expression, s = s 0 cos-^r. If 

we are considering a motion which is a resultant of two or more 
super-imposed simple harmonic motions and if we select t = 0 
when e = 0 for one motion, in general e will not be zero for the 
other motion and we cannot then avoid taking account of an 
initial phase angle for at least one of the motions. 

If we super-impose two simple harmonic motions of the same 
amplitude, s 0 , the resultant displacement at any instant is the 


16 


MECHANICS 


algebraic sum of the two, thus for 

Si = So cos 27mit and S 2 = So cos (27rn 2 t + e), 

(writing frequency, n, for 1 /T. 

The sum is 

s = Si + s 2 = s 0 cos 27mit + s Q cos (27rn 2 t + e) 

Which may be written 

s = 2s 0 cos y £ 2tt (m + n 2 )t+ej cosJ- JW(ni-n 2 )t-eJ 23 

This equation has many applications and is of particular im¬ 
portance in the study of electric oscillations, especially in con¬ 
nection with telegraphy and telephony both wire and wireless. 

19. Expressions for Acceleration, Period, Force. By two 
differentiations of equation 22 a or 22 b we find for acceler¬ 
ation the equation 

47r 2 0 

a = - -Tp- s 24 


and the general expression for the period of a simple harmonic 
motion. 

T = 27 t V-s /a 25 

The corresponding force equation for a simple harmonic motion 
is 


F=ma = 


47r 2 m 


s 


26 


PROBLEMS 

1. In the case of a stretched or compressed spring the return 
force is known to be proportional to the distortion and in the 
opposite direction, that is F = -cs where (c) is known as the 
spring constant. Define (c) in physical terms. Show that if 
a mass (m) is attached to the spring it will vibrate with a period 
expressed by the equation 

T = 2t V m/c 27 

2. Find the expression for the period of a simple gravity 
pendulum. What is the effect of the mass on the period in this 
case? 

3. A body has an acceleration of a =—10 s cm./sec 2 . Its 
maximum displacement is 50 cm. Find the period, the maximum 
velocity, the maximum acceleration. 

4. Show that the projection of a uniform circular motion is 
simple harmonic. 

5. A mass of 50 gms. is hung on the end of a spiral spring. 






WORK AND ENERGY 


17 


When set vibrating it is found to have a period of 4 sec. Find 
the “constant” of the spring. 

6. A mass of 1 kg. is 
supported on frictionless 
wheels and is held on the 
inclined plane by a spiral 
spring. When hung freely 
a weight of 100 gms. 
stretches the spring 2 cm. 
Find the period with which 
the 1 kg. will vibrate when 
displaced. 

7. A stick 2 cm. square and 100 cm. long is loaded with lead on 
one end so as to float upright in water. Mass of whole stick 
plus lead is 250 grams. When lowered or raised from its equilibri¬ 
um position it will “bob” up and down. Neglecting friction, 
find the period. 




8. Show that a vibrating column of liquid 
in a U-tube has simple harmonic motion. 


Fig. 11 


9. A body of mass (m) vibrates with S. H. M. in a straight 
ine. Find the average displacement during a half vibration 
(starting from the rest position). 


WORK AND ENERGY 


20. Work is defined as the product of force times the distance 
traveled in the direction of the force during the time the force 
is acting. For a variable force we may consider that through 
an infinitesimal distance the force may be considered as devi¬ 
ating from a given value F by but an infinitesimal amount and 
that therefore the work dw is 

dw = Fds 28 


This may be written 


dw 

ds 


= F 


which says, mathematically, that work is 
rate of change with distance equals force. 


the quantity whose 
To find w we must 








18 


MECHANICS 


find the quantity which when differentiated with respect to s 
gives F. This quantity is called the integral of F with respect 
to s. This is written mathematically as 

w = J*Fds 29 

The c.g.s. unit of work is the work done by a force of 
one dyne acting through a distance of one centimeter. This 
unit is called the erg. The joule is 10 7 ergs. 

In the English system the unit of work is the foot-pound , the 
work done by a force of one pound acting through a distance of 
one foot. 

21. Power, P is the time rate of doing work. In mathe¬ 
matical language 

P = dw/dt 30 

Hence w = J* Pdt 

Also by combination of 28 and 30 we find 

P = F 4r-= Fv 31 


The c.g.s. unit of power is the erg per second. A more con¬ 
venient unit is the watt which is the joule per second. Still 
another unit is the kilowatt (1,000 watts). A common unit of 
work in measuring electrical energy is the kilowatt-hour. It is 
the work done (or energy consumed) in one hour when the rate 
is a kilowatt (1,000 joules every second). 

The English unit of power is the horse-power , which is 550 
foot-pounds per second. 

22. The Graphical Representation of Work. If force, F, is 



5, ds \ 


Fig. 13 


plotted against distance, 
s, it may be shown that 
the area under the curve 
on such a diagram gives 
work. Suppose the force 
has the constant value, F, 
(Fig. .12), and the body 
travels a distance, Si, in 
the direction of the force, 
then work, being F x Si, 
is obviously the area udner 
the force-distance line. 
If the force is variable 
(Figure 13) and we wish 
to find the work when the 
body moves from Si to 
S 2 , it will now be neces¬ 
sary to divide the distance 
into infinitesimal portions 












WORK AND ENERGY 


19 


of length, ds, through which the force varies from some value, 
F, by only an infinitesimal amount. In such an infinitesimal 
range, the work is dw and dw = Fds. The summation of all the 
strips obviously gives the total work between the limits, Si and 
S 2 , and this is the area under the force distance line. We have 
seen, equation 29, that total work in the case of a variable force 
is 

w = J Fds 

Hence the process of integration is here a process of summation- 
23. Illustration. The Stretched Spring. Let us consider as 
an example of this process the work done in stretching a spring. 
The force, F, applied must at each instant equal the elastic 

return force, F' ( =—cx), 
and be in the opposite 
direction. The applied 
force is therefore F = + cs, 
s being the distance the 
spring is stretched at any 
instant. Substituting this 
value in equation 29 we 
have 

32 

Placing the constant outside the integral sign we have 
w = c f sds 

The integral of this (the quantity which when differentiated 
with respect to s gives s) is 



That is, w = c (-— + c') 33 

where c' is the constant of integration. If we start measuring 
work when the displacement is zero, that is, take w = 0 when 
s = 0, then c' = 0. Equation 33 then becomes 



Inspection of Fig. 9 shows that the summation of all the strips 
is the area of the triangle, which is 

1 1 1 2 
- SiFi or - Si. csi = - csj 

Hence the value of s in equation 33 is the final value, Si. 

In general we might wish to find the work done in stretching 



w 


= f csds 







20 


MECHANICS 


the spring from the values So to si. In this case we start measur¬ 
ing work when the displacement is s 0 , that is, take w = 0 when 
s = s 0 . Then c', equation 33, becomes 



Hence work in stretching from So to Si is 


csf 

w = _i — 


34 


Inspection of Fig. 14 shows that yi cs^ is the work which had pre¬ 
viously been done in stretching to So- % cs| is the total work in 

stretching to Si. The difference between these is the work done in 
stretching from s 0 to Si. The integration in this case is said to be 
between the limits So and Si and the expression is written as 



If we were to consider the work done by the elastic return force 
of the spring F' ( = — cx) we would have 


w = 




35 


The work is negative since the distance traveled in the direction 
of the force is negative. 

24. Energy. Energy may be defined as the capacity to do 
work. When a force acts upon a body and does work upon it, 
(1), the momentum of the body may be changed, (2), the body may 
be moved in opposition to return or restoring forces, (3), the body 
may be moved in opposition to fricitional forces which do not tend 
to return the body to its initial position. In the first case the 
body may give up its increased momentum when it encounters 
opposition to its motion and mav therefore exert a force (equal 
to the rate of change of momentum) and do work upon an opposing 
body. The energy a body possesses by virtue of its motion is 
called kinetic energy. In the second case when work is done 
against restoring forces, these restoring forces, when the applied 
force is removed, will return the body to its initial position and 
thus do work equal to that done upon the body. In this case the 
ability to do work is potential. This energy is stored in the system. 
A body possessing energy by virtue of its position or the configur¬ 
ation of its parts is said to have potential energy. The stretched 
spring has potential energy. A body lifted against gravity to a 


WORK AND ENERGY 


21 


certain height has potential energy of position. It possesses the 
ability to exert force due to the pull of gravity upon it and do work 
upon opposing bodies. In the third case, while work is done upon 
the body the body does not acquire the capacity to do work. 
The energy is not stored in the body either as kinetic or potential 
but has been dissipated as heat. 

Experiment shows that the energy acquired by a body (if 
none has been dissipated as heat) is equal to the work done upon 
it. When energy is acquired by one body experiment shows 
that an equal amount must have been lost by the body through 
which the force was exerted. Energy may be converted from one 
form to another but experiment shows it cannot be created or 
destroyed. This statement is known as the Law of the Conser¬ 
vation of Energy. For example, imagine a moving body suddenly 
attached to the end of a spring, the spring will be stretched and the 
body brought to rest. We have here the kinetic energy of the 
moving body converted to potential energy of the stretched 
spring. Consider a projectile fired vertically with a certain 
initial velocity (v), after a certain time the upward velocity will 
be reduced to zero. Gravity has been doing work upon the body. 
At the instant the velocity is zero the body will have potential 
energy equal to the kinetic energy lost. 

The kinetic energy of a moving body is given by the expression 

E k =-^-k mv 2 36 

This expression is derived in the following way. The work done 
upon a body is 

w = j Fds. Since F = kma we have 

w = f kmads. From the equations a = — an H 
J dt 

ds 

v = — ,by eliminating dt, we have ads = vdv. Substituting this 

we have w = kmj" vdv = kmv 2 

v is the velocity acquired under the influence of the force, F. 
The expression^ kmv 2 is equal to the work done upon the body. 
In coming to rest the body may do an amount of work equal to this, 
hence this expression represents the kinetic energy of the body. 

As a further illustration consider the case of a body at a height, 
h, above the ground. Its gravitational potential energy is 
(kmgh) and if allowed to fall freely it will acquire kinetic energy 
equal to J^kmv 2 and these must be equal. This is shown as 


22 


MECHANICS 


follows. The expression (kmgh) is the expression for work 
done upont he body, m, against the force of gravity (kmg) 
through the distance, h, and therefore the potential energy 
E p = kmgh. When the body falls through the distance, h, the 
work done by gravity is (kmgh).. The acceleration, g, being 
uniform, 

h = -i- gt 2 and v =gt 

Making these substitutions in the expression for work (kmgh). 
we have 

kmgh =ykmv 2 

PROBLEMS 

1. If one pushes with a constant force on a body which is 
moving initially at the rate of x feet per second, will more or less 
work be done during the first second than when the body is moving 
initially at the rate of 2x feet per second? If so how much? 
How much work done in 10 seconds in each case? 

2. A force equal to the weight of a kilogram acts on a body 
continuously for 10 sec., causing it to move in that time a distance 
of 10 m. Find the mass of the body. 

3. A body weighing 20 pounds slides 100 feet down a 10 per¬ 
cent incline and acquires a velocity of 20 feet per second. How 
much energy has been converted to heat through friction? 

4. Water flows into a mine 500 ft. deep at the rate of 100 ft. 
per second. Find the H. P. of a pump to be used to keep the mine 
dry. 

5. Show that in the case of a simple pendulum, the potential 
energy at the end of the swing is equal to the kinetic energy at the 
mid-position. Use integration. 

6. A ton of coal is hauled in a car of 400 lbs. out of a mine 
200 ft. deep. During the first half of the journey it is accelerated 
4 ft. /sec. 2 , and during the last half, retarded by the same amount. 
Find the forces exerted, the work done, and the average power 
exerted. 

7. A pile driver (50 kg.) falls 5 meters and drives a pile down 
10 cm. Find the work done, the average force exerted, the time 
it is acting, and the power exerted. 

8. A 5 grm. bullet is fired with a velocity of 300 meters /sec. 
from a gun that is clamped in a vise that is supported from the 
ceiling by cords 2 meters long. Total mass of the gun and vise, 


ANGULAR MOTION 23 

10 kg. Find the arc through which the ballistic pendulum will swing. 

9. The above mentioned bullet strikes a lead target at the other 
end of the factory. The target is supported in the same manner 
as the vise, has a mass of 2 kg. and swings through an arc of 30 
cms. Find the target velocity of the bullet. 

10. A long chain having a length (1) and a mass per unit length 
(m) is attached by one end to a windlass and suspended in a shaft. 
Find the work done in winding up yi of its length. 

11. In a steam engine, the average pressure of steam during a 
stroke is 160 lbs. per square inch. The length of stroke is 32 
inches and the diameter of the piston is 24 inches. If the engine 
makes 60 revolutions per minute, find its horse-power. (Pressure 
= force /sq. in.) 


ANGULAR MOTION 


25. General Equations. If a rigid body is set to rotating about 
some axis the angle turned through from some initial position 
is called the angular displacement, 4>. The rate of change of 4> 
is the angular velocity, co, and the rate of change of co is the 
angular acceleration, a. In mathematical terms 


d<£ 

dt 

dto 

dt 


d 2 $ 

dt 2 


37 


38 


From these equations we also have 

$=/ codt 39 

a ) = J'adt 40 

26. For Uniformly Accelerated Angular Motion reasoning 
exactly as in the linear case, we may establish the equations 

co = at + <F 0 41 


$ = | at 2 + <o 0 t + 3> 0 


42 


27. For Angular Simple Harmonic Motion, reasoning exactly 
as in the linear case and starting with the equation of angular 
simple harmonic motion as 


d 2 4> 

dt 2 


= -k$ 


43 



24 MECHANICS 

we may establish the corresponding equations as the solutions, 
namely. 

<f> = <i> o gin ( vTt +e-) 44 

Interpreting as before we find k = 47r 2 /T 2 , whence 44 becomes 

= $>o sin t + e) 45 

and from 43 the expression for the period of an angular simple 
harmonic motion 


T = 2x V-Q/a 4b 

28. Vector Representation of Angular Quantities. If the 

distance, s, along the arc of a circle subtended by a given angle, 
4>, is divided by the radius of the arc, the quotient is the number 
of radians in the angle. That is 

, s 
4 > = - 
r 

The direction of s is at any instant tangential to the arc, and r 
is at right angles to it. The quotient of two mutually perpen¬ 
dicular vectors is a third vector at right angles to each of the two. 
Hence the angle is represented by a vector along the axis of ro¬ 
tation. We may thus represent angular displacements, velocities 
and accelerations and they may be compounded or resolved as in 
the linear cases. 

29. Moment of Force. The effectiveness (or moment) of a 
force in changing the angular velocity of a body depends not only 
upon the magnitude of the force but upon the distance of the 
point of application of the force from the axis. Consider the 

case of the plank balanced 
as shown, (Fig. 15), with 
weights Wi and w 2 . If d 2 is 
twice as great as di then 
w 2 need only be half as 
great as wi to balance Wi. 

Each force (wi and w 2 ) tends to produce rotation about o, but 
in opposite directions. The effectiveness or moment of each force 
about o is the same and depends only upon the magnitude of each 
force and its distances from o. The product of force and lever 
arm (measuring as it does the effectiveness to produce rotation) is 
called the moment of force. In case of equilibrium the sum of all 
the moments about any axis is zero. 



Fig. 15 






CENTER OF MASS 


25 


PROBLEMS 

1. At the instant power is cut off from a motor, the armature is 
going 2200 r.p.m. It comes to rest in 10 sec. Find (a) acceler¬ 
ation; (b) total number of revolutions before stopping; (c), the 
distance it would have travelled if it had been rolling on the ground 
during the above time. Its diameter is 20 cm. 

2. A cylinder, 1 foot radius, rolls down an inclined plane 10 
meters long in 8 seconds. Calculate the angular acceleration, 
and the angular velocity at the bottom of the plane. 

3. Find expressions for co and $ for motion in which a = 0 


CENTER OF MASS 


30. The center of mass of a body is defined as that point, located 
usually within the body, at which a force, applied in any direction, 
will give the body a motion of translation only, i.e., there will be 
no rotation, and all particles of the body will move with the same 
acceleration. 


< - - - 


V 


Let us consider the ideal 
simple case of two particles, 
mi, and, m 2 , held rigidly 
together by a massless con¬ 
nection as shown. We may 


Fig. 16 

find an expression for the position of the center of mass in the 
following way. Assume the origin of axes of reference at some 
point, P, and the center of mass of the system at point, O. Re¬ 
quired, an expression for the distance, D, the distance of the center 
of mass from the origin. 


By definition a force applied at O produces translation only. 
The sum of the moments about any point must be zero if there 
is to be no rotation. We may therefore equate to zero the sum of 
the moments about P. These moments arise from the applied 
force, F, and the reactions offered by mi and m 2 . By Newton’s 
Second Law 


F = (mi + m 2 ) a 

We may therefore write for the moments about P, 
(mi + m 2 ) aD — miaxi - m 2 ax 2 = 0 


whence D = 


mixi + m 2 x 2 
mi + m 2 


47 


31. We can extend the above reasoning to a rigid body. Let 






26 


MECHANICS 



the origin of co-ordinates of 
a rectangular system be at 
P. To find the x-co-ordi- 
nate of the center of mass. 
Let O be the center of 
mass. Apply a force at O 
parallel to the yz plane. 
Let a be the acceleration 
of an element of mass dm. 
The force acting on dm is 
dF = k*dnra. If the x- 
coordinate of dm is x and 
of 0 is X then the moment 
of this reaction force about 


O is k dm a (x-X). If no turning is to result the sum of all these 
moments for the whole body is zero, i.e., 

S k- dm- a (x-X) = 0 48 

Jx'dxn = /x dm = XM 

where M is the mass of the whole body. Therefore 

X = Jxdm/M 49 

(xdm) is called the moment of the mass (dm). Similar express¬ 
ions may be obtained for y and z co-ordinates of the center of 
mass. 


32. Velocity and Acceleration of Center of Mass. If equation 
is written in the form 


v _xidmi , x 2 dm 2 , 
X M + M + 


50 


we may differentiate this expression with regard to time, term by 
term, obtaining 

dX _ _ Vidmi 

dt~~ “ M 

and again 

d 2 X_ . _ aidmi 

dt 2 M 

Equations 51 and 52 give us the velocity and acceleration of 
the center of mass in terms of the velocities and accelerations of 
the individual particles which make up the mass. 

Writing equation 52 in the form 


, v 2 dm 2 . 
' h M + 


51 


, a 2 dm 2 
M 


52 










27 


CENTER OF MASS 


a _ aidmi -f- a^dm 2 + — Sadm 

M M~ 53 

certain peculiar properties of the center of mass may be noted, 
Thus (aidmi + a 2 dm 2 + • • •) represents the sum of the reactions 
of all the particles which make up the mass. By Newton’s Third 
Law, 2f = Sadm, where 2f is the sum of the externally applied 
forces. We may therefore write 53 as 



54 


From 54 it follows that if the sum of all. the externally applied 
forces is zero, the acceleration of the center of mass is zero. 

33. Action of a Couple. A couple is composed of two equal and 
oppositely directed, parallel, but not concurrent forces. If a 
couple acts upon a body, the sum of the applied forces in any 
direction is zero; it therefore follows, equation 54, that the 
acceleration of the center of mass is zero. Therefore we must 

conclude that the rotation 
produced by a couple no 
matter where applied is always 
about an axis through the 
center of mass. As an ex¬ 
ample consider the rod (Fig¬ 
ure 18) with center of mass 
at 0 and let the same couple 
be applied in the different 
positions as shown. If the couple is simply transferred parallel 
to itself, exactly the same effect will be produced in each case, 
i. e. exactly the same rotation about the same axis through 0. 
(If the plane in which the couple acts is changed then the rotation 
would be about a different axis but one which must passthrough O). 

34. Motion of a Rigid Body when the Resultant of the Forces 
acting upon it is not Zero and does not pass through the Center 

of Mass. Let F (Figure 19) 
be the resultant of forces 
applied to the body whose 
center of mass is at 0. 
Without effecting the motion, 
we may apply at 0, equal 
and opposite forces, Fj, and 
F 2 , each equal in magnitude 
to F and parallel to it. 
By so doing we are enabled to 
resolve the motion into one 
of pure rotation about the 
center of mass , 0, pro- 













28 MECHANICS 

duced by the couple (F, F 2 ) and one of pure translation produced 
by the force (Fi) acting through the center of mass. 


35. Center of Gravity. 



center of gravity of a body is 
the point through which the 
resultant of all the gravity 
forces acting on the particles 
of the body passes. We may 
consider the weight of a body as 
the sum of the gravity pulls on 
all the particles of the body. 
Since the body falls with trans- 
latory motion only, the resul¬ 
tant of all these gravity forces 
must act through the center of 
mass. Center of gravity is 
therefore coincident with center 
of mass. 


PROBLEMS 

1. Using equation 49 find the distance from one end of the 
center of mass of a uniform rod length L, radius r, density 8. 

2. Find the center of mass of an isosceles triangle of tin, base b, 
height h. 

3. Find the center of mass of a cone of height h, density d, 
(uniform) and radius of base r. 

4. The density of air at the earth’s surface is 0.00129 grams per 
cc. Three kilometers above the surface it is 0.00080 grams per cc. 
Find the center of mass of a column of air 1 sq. meter in cross 
section and 3 km. high. 

5. A rectangular board, Fig. 21, weighing 10 kg. rests on 
universal roller bearings on a horizontal smooth surface, (i.e., it is 
free to move without friction in any direction). A force F, of 3 kg. 
wt. is applied to the body for 0.1 sec. In what direction will the 
body move, and what will be the velocity of its center of mass? 
What other motion will it have? 

6. A bar (Fig. 22) is dropped from an aeroplane traveling north¬ 
ward with velocity, v. It receives a horizontal blow, F, (in 
the nature of a kick southward) just as it leaves the machine. 
Discuss the motion as to nature and direction. Discuss the motion 
if the impulse of F, (1) equals mv, (2) is greater than mv, (3) is 
less than mv. 



























MOMENT OF INERTIA 


29 



Fig. 22 


MOMENT OF INERTIA 

36. Any attempt to change the angular motion of a body meets 
with opposition due to inertia just as any attempt to change 
the linear motion of a body meets with like opposition. The 
opposition in the angular case depends not only upon the mass 
but upon the way the mass is distributed about the axis of rotation. 
The factor which determines the opposition to a charge in the 
angular motion of a body is called the rotational inertia or 
moment of inertia of the body. The moment of inertia is 
numerically equal to the opposition per unit angular acceleration 
and is given by the expression 

I = k j>dm 55 

where dm is an element of mass, r the distance of thie element 

from the axis. This ex¬ 
pression follows from the 
definition of moment of 
inertia as may be shown 
by the following reasoning. 
Let the body (Fig. 23) be 
mounted on a shaft (0) 
and consider the reactions 
when the shaft starts rotat¬ 
ing causing acceleration in 
the direction of the arrow. 
If a is the angular acceler¬ 
ation, the linear acceleration of the mass dm is a = ar. The force 
by which the adjacent parts of the body must act on dm is 
dF = kdma = kdm a r 

This represents also the reaction of the element dm. This reaction 
exerts a moment of force, dL ( = rdF), opposed to the motion, and 

dL = kr 2 dm a 



Fig. 23 







30 


MECHANICS 


The sum of the reacting moments due to all the mass elements 
must by Newton’s Third Law (extended to the angular case) 
equal the moment of force or turning torque , L, applied through 
the shaft, that is, 



It will be noted from this expression that k ( r 2 dm is equal to the 


opposition per unit acceleration and therefore by definition is the 
moment of inertia. For any absolute system of units, i.e. a 
system in which k = 1, we have 



56 


It follows from the above that the general expression for 
torque is 

L = k I a 57 

Equation 57 is the fundamental equation of angular motion. 
It is the expression of Newton’s Second Law as applied to angular 
motion and is the angular analogue of Equation 15. 

By comparing Eq. 57 with Eq. 15, F = kma, we may note 
that torque, L, is analogous to force, F; I, moment of inertia 
is analogous to mass, m; a, angular acceleration is analogous 
to linear acceleration, a. 

37. Angular Momentum, (sometimes called Moment of 
Momentum). The product Ico, where co is the angular velocity, 
is called the angular momentum. This quantity is strictly analog¬ 
ous to linear momentum, mv. 

Impulse of a Torque. The product of torque times time during 
which it acts, Lt, is called the impulse of a torque. It follows from 
equation 57 that impulse is proportional to angular momentum 
acquired under the action of the torque, that is 

Lt = k Ico 58 

38. Work. In turning a body through an angle $ work is 
given by the expression 



59 


w 


and is expressed in the same units as in the case of translation. 
We may define work in the angular case by analogy with the linear 
case, whence equation. 59 or we may show independently that 
equation 59 expresses work. Thus, L is equal to an applied 
moment of force, Fr, and d<£ = ds/r-where ds is a linear distance 
along an arc. Substituting these in 59 we have 



Therefore 59 is an expression for work. 



MOMENT OF INERTIA 


31 


39. Energy. Angular Kinetic Energy is given by the expression 

E = 1 k I&> 2 60 

K 2 

This is obtained by applying the same reasoning to the angular 
case as in the case of translation. Likewise potential energy , 
energy stored by virtue of work done against a restoring torque, 
such as energy stored in a twisted shaft, is given by the equation 

E = k L $ 61 

p 

40. Moments of Inertia about Parallel Axes. If I 0 is the 

moment of inertia about an axis through the * center of mass, 
then the moment of inertia; I a , about some other parallel axis is 

62 


k = !« 


+ MhV 


where (h) is the distance between axes. This is shown as follows: 

/ Imagine an axis through 

Q the center of mass, o, 

/ (Fig. 24) and a parallel 

axis through a, distant, 
h, from - o. Take any 
element of mass , dm, dis¬ 
tant, r 0 , from o, and, t%, 
from a. By trigonometry 
2 2 2 

r =r -f h —2r r cos 0 
10 1 o 

multiplying each term by 
dm and integrating we 
have 


Jr" dm = j'/dm+ h 2 Jdm— 2 r 0 cos 6 - dm 

Draw rectangular co-ordinates through o, letting the x-axis 
coincide with h. It is then apparent that Jr o cos0 ’dm is the 

siim of the mass moments about the center of mass which by 
equation 49 is zero. Therefore we have 
2 2 .2 /• 

Jr dm = Jr^dm + h J dm 

which may be written I = Io +Mh 2 






32 


MECHANICS 


41. Motion of a Rolling Body. Proposition. In the case of a 
rolling disk or sphere we may consider the motion as a, translation 
of the center of mass and rotation about an axis through the 
c. of m. with angular velocity, co, or (b), as pure rotation with 
the same angular velocity, co, about an instantaneous axis tangential 
to the disk or sphere and in the surface upon which the body 
is rolling. 

It was shown in the treatment of center of mass that a force 
acting on a rigid body produces in general translation of the 
center of mass and rotation about it, the only exception being 

when the force acts through 
the c. of m., in which case 
it produces translation only. 
We may consider the motion 
which a body has acquired, 
to be the result of force 
action at some time or 
other, and therefore it fol¬ 
lows that any motion con¬ 
sists of pure translation 
and of rotation about the 
c. of m. It remains to be 
shown then that the mo¬ 
tion of a rolling body may 
also be considered as pure rotating about an instantaneous axis 
with the same velocity as about the c. of m. 

Consider an element of the body at point, x, (Figure 25). 
It has horizontal velocity, v x , equal to that of the c. of m. (as 

has every particle of the body) and velocity, v 0 , in a circular path 
about the c. of m. Its resultant velocity is v. v 0 is perpen¬ 
dicular to r 0 , and v x is perpendicular to r x . Therefore the two parallel¬ 
ograms whose sides are respectively the velocities and radii indicat¬ 
ed, are similar, and therefore the triangles whose sides are respective¬ 
ly r, r Q , r x , and v, v 0 , v x are similar. Therefore, since v 0 is perpen¬ 
dicular to ro and v x to r x , it follows that v is perpendicular to r. 
That is, the resultant velocity is in the direction of rotation 
about the point, a. From the similar triangles we have also 

v Vo - . v - v 0 

— = — and since •— = co a and — = co 0 

r r 0 r r 0 

it follows that co a — co 0 

42. Equation of Motion in the Case of Rigid Bodies. In 

general we may set up the necessary equations either from the 
point of view of torque relations or that of energy relations. 






MOMENT OF INERTIA 33 

As an example let us con¬ 
sider the motion of a disk 
rolling down an incline. 

(1) Considering torque re¬ 
lations, we have 

mgr sin 0 = I a a 63 

(2) Considering energy re¬ 
lations we have 

mgh = ~ I a w 2 64 

where go is the angular velocity acquired when the disk has reached 
the foot of the incline. 

We may obtain 64 from 63 in the following way. From the 
laws of uniformly accelerated motion (assuming the disk starts 
from rest) 

a = —^jr . Also 4> =— and sin 0 = — 

24> r s 

Where (s) is the length of the plane, h, the vertical distance. 
Substituting these values in 63 we get 64. Or we may write 
64 as 

mgh = loco 2 + y mv2 > 65 

by simply applying the relations, 

v 

I a = Io + mr 2 and go' = — 

Eq. 65 might be written directly as a result of the proposition 
proven in paragraph 41. 



Fig. 26 


PROBLEMS 

1. Using the calculus method, compute the following moments 
of inertia: 

a) A uniform thin rod about an axis perpendicular to the rod 
and passing (1) through the middle of the rod and (2) through 
one end of the rod. 

b) A disk about an axis through center and (1) perpendicular 
to disk: (2) parallel (i. e. in plane of disk). 

c) A cylinder about a transverse axis a distance (r) from 
the center of gravity. 

d) A rectangular disk about an axis through center and 
(1) bisecting sides of length a; (2) bisecting sides of length 
b; (3) perpendicular to disk. 




34 MECHANICS 

2. A cylinderjof iron, den¬ 
sity 7.8 grams per cc. 50 
cm. long and 10 cm. radius 
is mounted so} that it can 
rotate about its axis of 
figure, a a. 

a) Compute its moment of inertia. 

b) What angular acceleration must it have in’order to acquire 
an angular velocity of 50 radians per second in 3 seconds? 

c) What would be its kinetic energy when going at this rate ? 
What power was applied to the cylinder during the 3 seconds ? 

d) What torque exclusive of friction must act on the cylinder 

to produce the avbove acceleration? -fj 

e) When going at the above rate, power is removed and the 
cylinder comes to rest in 1 minute. What is the frictional 
torque? What was the total torque acting on the cylinder 
while it was speeding up ? 

f) A cord is wound several times around the cylinder and a 
mass of 5 kg. is hung on the free end of the cord. How long will 
it take the 5 kg. to drop 10 ft. and what will be 

1) the linear velocity of the 5 kg., 

2) the angular velocity of the cylinder, 

3) the kinetic energy of each? 

3. Given three masses: 

a) A sphere of iron, 
mass 100 kg. 

b) A cylinder of iron, 

Fig. 28 mass 100 kg. and of same 

radius as the sphere 

c) A cube of iron, mass 100 kg. mounted on light frictionless 
wheels. The three masses are started at the same instant 
down the inclined plane. 

1) which will reach the bottom first? 

2) what will be the time of descent of each ? 

3) what will be the total kinetic energy of each and how 
will this energy be distributed between energy of translation 
and of rotation? Neglect friction. 

Density of iron = 7.8 grams per c.c.) 

4. Derive the equation for the compound physical pendulum. 

T = 2tt\/ I/mgh 

5. An iron sphere is suspended about an axis tangent to its 
surface. Radius of sphere R. Find its period. 

6. A bar loaded with lead at one end and fitted with a pair of 
adjustable knife edges is called a Kater’s pendulum. It is a 










GYROSCOPIC MOTION 35 

reversible pendulum, that is the bar may be set to vibrating a- 
bout one set of knife edges or inverted and set to vibrating about 
the other set. Show that the length of a simple pendulum adjusted 
to vibrate synchronously with the bar is equal to the distance 
between knife edges, when the period is the same about either. 
Derive the expression for (g), (1) when the periods are not exactly 
the same for each knife edge, (2) when the periods are exactly the 
same. 

7. The disk shown 
(moment of inertia 10,000 
gm. x. cm. 2 ) is given 
an initial angular velocity 
counter clock-wise of 5 
R. P. S. and the suspend¬ 
ed mass (2000 gm.) a 
corresponding linear vel¬ 
ocity. Find height to 
which the mass will as¬ 
cend before coming to 
rest. 

8 . Moment of inertia 
of a body is 7000 gm., 
cm. 2 If the torque acting 
is 35000 dynes, cm., find 

the angular velocity at the end of 8 seconds if the initial velocity 
is 12 radians per sec. What will be the change in kinetic energy? 

9 . A certain flywheel has a velocity of 100 R. P. M. A torque 
is applied and at the end of three seconds its velocity is 300 R. P. M. 
The power applied is uniform and equal to 3 kilowatts. Compute 
(a) the moment of inertia, (b) the torque applied when the velocity 
is 300 R. P. M. 

10. A horizontal drum two meters in diameter is on the same 
shaft as a heavy flywheel. The flywheel is 3 meters in diameter, 
and has a mass of 5000 kg. which may be considered as concentrat¬ 
ed in the rim. A cage whose mass is 1000 kg. is attached to a 
rope passing over the drum. At the instant the power is cut 
off the drum is making 20 revolutions a minute. Find how far 
the cage will rise before coming to rest, neglecting friction and 
mass of drum. 



GYROSCOPIC MOTION 

43 . Gyroscopic Motion is the angular analogue of Uniform 
Circular Motion. In u. c. m. the rate of change of 



36 


MECHANICS 



momentum 


. mv 2 v 

is — or m — 
r vr 


Fig. 30 


which is mvco or mv ,. . 

at 

This is shown graphically 
by means of the vectors 
in Fig. 30. In the vector diagram, (figure to the right), the vector, 
Vi, plus the distance along the arc, vd<I>, equals the vector, v 2 , or the 
change in v is vd<3>. Hence the rate of change of momentum is 

d$ d$ 

mv -r— and the force is F = kmv -j-. 
dt dt 

The Angular Case. Consider a body, M, spinning with velocity 





U) 




Fig. 32 


co about a horizontal axis 
OA. (Fig. 31). Suppose 
the axis OA is rotating 
(precessing) counter clock¬ 
wise in a horizontal plane 
about the axis OB. The 
rate of change of angular 
d<i> 

velocity is co ^ (Fig. 32). 

The product of these two 
vectors is a third vector 
at right angles always to 
OA and OB. At the in¬ 
stant shown this axis pro¬ 
jects out from the paper 
toward the reader. 

The corresponding torque 
d3> 

is klco - 7 - This torque 


must equal the applied moment of force which is MgR. It will 
be noted that this applied moment is made up of two vectors 
at right angles and that the angle about which it is exerted agrees 
with the axis of precession as above discussed. Therefore the 

d<£ 

equation of this motion MgR = klco — = klcoco' 66 

d<£. 

where co' = -^-‘,the velocity of the precession about the axis OB. 


PROBLEMS 

1. A boy swings about his head, in a circle 1 m. radius a bullet 
whose mass is 20 g. What is the pull on the string when the 
bullet is making 1 revolution per second? 













ELASTICITY 


37 


2 . An engine flywheel is spinning about a horizontal axis 
with an angular momentum of 50,000,000 c.g.s. units. How 
great a torque will be required to rotate the axle of the fly-wheel 
(still spinning) about a vertical axis at the rate of yi radian per 
sec. ? About what axis must this torque be exerted? 

3. A high speed engine with its shaft athwart ship makes 240 
r.p.m. The rim of the fly-wheel has a radius of 3 ft. and a mass of 
600 lbs. Calculate the moment of inertia of the wheel (rim). 
The maximum angular velocity attained by the vessel in rolling 
is 1 /10 radians per sec. Calculate the maximum torque acting 
on the fly-wheel shaft due to gyrosocpic action and specify how the 
torque acts. [Nichols and Franklin, Elements of Physics]. 

ELASTICITY 

44. An Elastic Body is one in which a distortion or strain is 
opposed by molecular forces which are proportional to the strain 
and tend to restore the body to its original shape or volume. 
The strain is the distortion per unit length or per unit volume 
or per unit area, depending on the character of the distortion. 
The stress is usually considered as the return force per unit area 
and is equal to the distorting force per unit area. 

Hooke’s Law is a statement of the experimental fact that stress 
is proportional to strain within what is known as the elastic limit. 

Longitudinal strain, —an extension or compression along the 
longitudinal axis of a body produced by a force acting along 
that axis. It is numerically the change in length per unit length. 
The corresponding stress is the internal force per unit area across 
a section normal to the axis. The stress times the area 
equals the distorting force. The ratio of longitudinal stress to 
longitudinal strain is called Young's Modulus ( M) 

M = F//ae 67 

Volume strain — A strain produced by external forces which 
are equal along the three axes. An example is the change in 
volume produced by hydrostatic pressure. The stresses are the 
same in the three directions and equal to the external pressure. 
The ratio of stress to strain in this case is called the bulk modulus 

03 ). 



Shearing strain or shear, — a strain produced by an extema- 
moment of force. The distortion (or strain) consists in a dis 


38 


MECHANICS 



placement of any 
given section past 
adjoining parallel 
sections. The shear 
is measured by the 
angle of displacement 
(<£>), Fig. 33. The 
stress is the elastic 
return force per unit 
area along any sec* 
tion and is tangen¬ 
tial to the section. 


The shear modulus, n, also called slide modulus or rigidity modulus, 
is the ratio of stress to strain in case of shear. 


n = 


_F 

a<£ 


69 


Hooke’s law is strictly true 
only for very small strains. 
Typical stress, strain re¬ 
lations for a sample of iron 
is shown in Fig. 34. Point 
(a) beyond which we have 
deviation from the strict 
proportionality between 
stress and strain is called 
th e elastic limit. When 
strained beyond point (a) 
the body does not come 

back to its original conditions but retains a slight permanent set. 
At point (b) a sudden yield to the distorting force occurs, after 
which a large permanent set remains, (b) is called the yield point. 
At (c) we have the breaking stress. In the case of longitudinal 
stress, the breaking stress is called the tensile strength. 

Area under a stress, strain curve may be shown to be equal to 
work. The work done per unit volume in straining a body to the 
elastic limit is known as the resilience of the substance. 

46. The Twisted Shaft or Wire. — Slide Modulus in Terms 
of the Constants of the Shaft or Wire. Fig. 36 represents a section 


45. Limitations of Hooke’s Law. 














ELASTICITY 


39 



of a twisted shaft or wire. <£is the angle of shear and varies for 
any r ng element from zero at the center to <£m at the surface. 
0 is the angle through which one end of the shaft or wire is twisted 
relative to the other. 


, s 

* = T 


hence 


» rd 

* = T 


6 = —and 
r 

Consider the force, df, on the shaded element whose area is r.da.dr. 
By definition of shear modulus, 

df = dil 
n r dr da $ r 2 dr da 6 


The turning moment of this force, df, is rdf = dL. 

1T n r 3 dr da 6 

dL -7- 

nd r R r 2w 

By integration L =- T - I rMr f da 
l J o J o 


Hence 


whence 


L = 


7r n r 4 ^ 
21 


70 


The applied torque, L, which is equal to this but opposite in direction 
is given by the expression L = —L o 0, where L 0 is defined as the 
constant of torsion of the shaft or wire, then 

2/Lo 


L„ = 


7T n R 


or n = 


71 


21 7TR 

47. If a weight is attached to a wire and the arrangement used 
as a torsion pendulum show that the period is 


T = 2 tt Vl/Lo 


72 


PROBLEMS 

1 . What must be the diameter of a steel wire 10 meters long if 
it is required to sustain a maximum load of 20 kilograms without 













40 


MECHANICS 


stretching more than one cm. ? Young’s Modulus for steel is 
20 x 10 dynes /sq. cm. 

2 . A is a steel wire (stretch modulus = 20 x 10 11 dynes/cm. 2 ) 
and B is a copper wire (mod = 10 x 10 11 dynes/cm. 2 ). Compare 
the stretches of the two wires, B being twice as long as A and 
having twice the cross sectional area. 

3. One end of a rod one meter long and 0.6 cm. in diameter is 
twisted through an angle of 0.2 radians. Where is the shearing 
strain the greatest, where is it the least ? Compute its maximum 
value. 

4. A steel shaft 40 feet long and 4 inches in diameter trans¬ 
mitting 100 horse power makes 14 r.p.s. What is the torque? 
If the slide modulus of steel is 12 x 10 6 pounds per square inch 
calculate the angle through which one end of the shaft is twisted, 
relative to the other end. 

5. (a) If the steel shaft in problem4 were to transmit 150 H.P. 
with the same torque at what speed must the shaft turn ? 

(b) If the shaft (problem 4) were to transmit 150 H.P. 
at the same speed what would be torque and through what angle 
would one end of the shaft be twisted relative to the other end? 

6 . The material of which a circular shaft is made will stand a 
maximum shearing strain of 0.01 without passing beyond the 
elastic limit. Through what angle can a shaft 50 ft. long and 1 in. 
in radius be turned without this strain being exceeded? Explain 
clearly. 

7. Explain how the torsion pendulum may be used to determine 
the moment of inertia of an irregular mass. 


HEAT 

THERMOMETRY 


48. Temperature Scales. Our first notions of temperature 
are obtained through our physiological sensations. We say 
that a body is at a “high” or a “low” temperature according to 
the wav it feels. Observation shows that the various physical 
properties of materials such as the length of a solid, the volume 
of a liquid, the pressure and volume of a gas, the electrical resistance 
of a metal, etc. change on heating. These changes have in various 
ways been adopted as measures of temperature. Since careful 
measurements have shown that these changes are not strictly 
proportional to each other, it has been customary to select some 
property arbitrarily and in terms of this property to define a 
temperature scale. The change of volume of mercury in glass 
affords a most convenient measure of temperature and scales 
chosen in terms of this property are known to everyone. 

A substance whose properties may be used as a measure of 
temperature is called a thermometric substance , the property 
chosen, a thermometric property. Any temperature measuring 
device is a thermometer. 

The fundamental temperature interval is the interval between 
the freezing and boiling points of water under standard conditions. 
The Centigrade degree is 1 /100 of this interval, the Fahrenheit 
degree 1 /180 of this interval. The temperature of melting ice, 
is assigned an arbitrary value—zero on the centigrade scale, 
32 on the Fahrenheit scale. 

If P 0 and Pioo are the values of the chosen thermometric 
property at the fixed points, the temperature on the Centigrade 
scale at which the property has the value P t is given by the equation 

= Pt-Po 73 

Pioo—P q 

100 

(This is the general expression for temperature on any Centigrade 
scale.) 

If we thus define temperature in terms of some particular 
thermometric property the change in some other property will 
not in general be proportional to temperature as thus defined. 


41 




42 


HEAT 


It may however be represented accurately by a series, 

Pt = Pc (1 + at + /3t 2 +.) 74 

in which P 0 is the value of the property at 0 °, Pt the value for 
temperature (t) and (a), (/3), ( 7 ), etc., constants depending 
upon the property and the temperature scale. 

49. Expansion. The ratio of the change in length per degree 
temperature change to the total length for any particular meterial 
is called the coefficient of expansion of that material. In general 
the relation between the length and temperature is expressed by 
an equation of the form of 74 and since the rate of change in length 
with temperature is the slope of the curve expressed by this equa¬ 
tion and since this slope varies with temperature it is apparent 
that the coefficient varies with temperature. The coefficient of 
expansion between 0 ° and t° is expressed (by definition) as 

a = \ /° , whence l t = U (1 + at) 74-a 

Iq t 

QUESTIONS AND PROBLEMS 

1 . Name various thermometric properties and describe how 
thermometers might be based upon each. 

2 . What sources of error can you think of in the mercurial 
thermometer. Explain why such a thermometer would not make 
a reliable standard. 

3. Adapt equation 73 to the Fahrenheit scale. 

4. Does the coefficient of expansion depend upon the units 
of length? does it depend upon the temperature scale? Explain. 

5. Show that the volume coefficient is three times the coefficient 
of linear expansion. 

6 . A thermometer with an arbitrary scale of equal parts 
reads 16.5 in melting ice and 158.5 in water boiling at standard 
pressure. Find the centigrade temperatures corresponding to 
8.6 and 124.0 on the thermometer. Ans.—5.2°C, 75.8°C. 

7. In terms of the mercury temperature scale, the resistance 
of platinum is given approximately by the expression 

R = Ro (1 + 0.00394t—0.00000059t 2 ) 

If we define temperature in terms of the change of resistance 
of platinum, what temperature on the platinum scale will cor¬ 
respond to 50°C. on the mercury scale? Ans. 50.4°. 

8 . A. compensated pendulum consists of an iron tube containing 
mercury. Its length is 108 cm. If the mass of iron is neglibile 
in comparison with that of mercury, compute the height of the 
mercury required for compensation. Coefficient of linear expan¬ 
sion for iron is 12 x 10' 6 per degree centigrade, coefficient of volume 
expansion of mercury is 18 x 10' 5 per degree centigrade. Ans. 
86.4 cm. 




EQUATION OF STATE 


43 


THE FUNDAMENTAL GAS LAWS AND THE 
EQAUTION OF STATE 

50. Boyle’s Law. If a gas is allowed to expand at constant 
temperature, experiment shows that as a first approximation the 
product, pv, remains constant. 

Charles’ Law. If temperature of a gas is allowed to change 
either pressure or volume must change, and the change in pressure 
or volume is proportional to the temperature change. This is 
an approximation and depends upon the temperature scale used. 

The Equation of State. If the pressure of a gas is used as 
the thermometric property, temperature on such a scale is given 
by the expression 

pt—Po 


t = 


Pioo—po 
100 


75 


pioo —Po 


=a where a is called the pressure coefficient. It is 

100 po 

the ratio of the change in pressure per degree to the pressure 
at the ice point. 

Therefore we may write 

j. _ Pt~Po 
ap 0 

If the volume of a gas is used as the thermometric property, temper¬ 
ature on such a scale is given by the expression 

v t — Vo 


and p t = po (1 + at) 


76 


t' =• 


and since 


Vioo—Vo 


100 v 0 


Vioo—Vo 
100 


= /3, the volume coefficint, we have 


77 


t' = 


Vt—Vo 
13 Vo 


and v t = v 0 (1 + j8t') 


78 


If the gas obeys Boyle’s law (then said to be a perfect gas) 
we may show that a = (3. and that t = t' or that the scales agree. 
Consider the gas beneath the piston in cylinder 1, Fig. 36, its condi¬ 
tion or state indicated by po, 

*11 


PaVoto 


m 


poK* 


mmm 


Pt V o 


v a t 


mi 


p Vt 


Vo, t 0 , (t 0 being taken 

as 0°C). Allow the tem¬ 
perature to change from 
t 0 to t', temperatures 
being expressed on the 
constant pressure scale, and 
allow the gas to expand at 
constant pressure to Vt. 


Fig. 36. 


































44 


HEAT 


By use of equation 78 we may write po Vt = po v 0 ( 1 + /3t) 
Next increase the pressure on the piston to pt, sufficient to 
bring the volume back to v 0 . By use of 76 we may write 
PtV 0 = poV 0 (1 + at) 

where temperature is measured on the constant volume scale. 
The change from the conditions in cylinder 2 to those in cylinder 
3 is an isothermal one and if Boyle’s law is obeyed then p 0 v t = ptVo 
and it follows that at = (it'. If the conditions of the experiment 
are adjusted so that the temperature indicated by t and t' is the 
boiling point of water, at which point the two scales agree, then 
t = t' and hence a = (5. Therefore at all other points t = t' 
or the two scales agree through out. 

We may next expand to any pressure and volume and if the 
temperature is kept constant 


Therefore 


PoV t = p t Vo = pv 
pv = poVo (l + at) 


Or 


whence 


pv = poVoa 




79 


where R = ^° v — and (v 0 /m) is the specific volume at 0°C and 

m 

atmospheric pressure. 

51. The Absolute Zero. In the discussion of the Kinetic 
Theory of Gases, which follows, pressure is shown to be due to 
the impacts of the molecules of the gas upon the containing walls, 
the molecules being in a state of rapid random motion, the temper¬ 
ature being a measure of their average kinetic energy. With this 
state of affairs when the velocity of the molecules becomes zero 
the pressure is zero and the temperature must be the absolute 
zero. In that case equation 79 gives the Centigrade temperature 
of the absolute zero, namely 



Hence if we add - degrees to the Centigrade temperature we 

have temperature on an absolute scale, that is, one in which the 
zero is absolute rather than arbitrary. Expressing temperatures 
on such a scale by the symbol, T, we have from eq. 79 

pv=mRT 81 

Equation 81 (or 79) is called the equation of state and of course 
applies strictly only to a perfect gas. 



45 


EQUATION OF STATE 

Hydrogen is very nearly a perfect gas as far as following Boyle’s 
law is concerned. On this account the scale of the constant 
volume hydrogen gas thermometer has been chosen by inter¬ 
national agreement as the standard temperature scale. Tempera¬ 
tures on this scale express very closely temperatures which would 
be given on the absolute or perfect gas scale. This thermometer 
gives us experimentally the nearest approximation we can get to 
absolute temperatures. 

52. Work on the (p, v) Diagram. When the piston, Fig 37, 

compresses the gas in the cylinder 
and moves down a distance 
ds work dW = Fds is done. 

F = pa, hence dW = pads but 
ads = dv. Therefore dW = pdv. 
On the pv diagram pdv is the 
area of the strip shown. Work 
done in compressing from volume 
V 2 to volume Vi is 

W-Jif 82 

If the curve AB is an isothermal 
(constant temperature line) we may intergrate equation 82 by 
expressing p in terms of v from equation 79. Thus 

m R (i +t ) 


Whence equation 82 becomes 

W=mR (|+t) J=mR ( ^ +0>°g ^ 83 

The fact that this comes out negative since V 2 >vi, means that 
work is done on the gas. If this gas were expanding the limits 
would be from Vi to V 2 and the work would be positive (external 
work done by the gas.) 

QUESTIONS AND PROBLEMS 

1. How should a gas thermometer be calibrated to read centi¬ 
grade temperature? 

2 . Show that the absolute temperatu re of zero degrees centigrade 
is (1 /a) where a is the coefficient of expansion of a perfect gas. 













46 


HEAT 


3. Find the work done when 5 grams of hydrogen at 0° C. 
and initially at atmospheric pressure expands without temperature 
change to twice the initial volume. What change in pressure 
occurs? (The density of hydrogen under standard conditions is 
0.00008988 gms. per cc. Ans 3846 x 10 7 ergs. 

4. Calculate the volume of a gram-molecule of hydrogen 
under standard conditions (a gram-molecule of a gas is its 
molecular weight in gms.) 

5 . A barometer tube is filled with mercury to within one 
inch of the top. After inversion the air expands and occupies 
12 inches of the tube and the mercury stands 27 inches above 
the trough. Find the true height of the barometer. Ans. 29.45 
inches. 

6 . At the temperature of water boiling at a pressure of 760 
mm. of mercury the hydrogen in a hydrogen constant volume 
thermometer bulb is at a pressure of 766 mm. When the bulb 
is placed in a bath of melting lead the pressure is increased to 
1232 mm. What is the temperature Centigrade of melting lead? 

7. Compute the work necessary to compress a gas originally 
having a volume of 20 litres and a pressure of 76 cm. of mercury 
to 10 litres, the temperature remaining constant. Express the 
result in ergs. Ans. 1053 ergs. 

THE KINETIC THEORY OF GASES 

53. Assumptions of the Kinetic Theory. Chemistry teaches 
that a finite amount of any substance is made up of an enormous 
number of very small discrete particles called molecules, which, 
for the same substance, are all alike in form, size, and mass. The 
Kinetic Theory of Gases further assumes that, in the case of a gas 
of not too great density, the actual bulk (volume) of the molecules 
is very small as compared with the volume occupied by the gas; 
that when the gas is in “equilibrium” the molecules are distributed 
with approximate uniformity throughout the container; that 
they are moving about in all directions with different individual 
velocities; and that when they bump into the walls of the container, 
they rebound without any loss of energy (i.e. with perfect elasticity) 
and thereby exert a pressure against the walls. But, although 
the individual velocities differ widely and the velocity of any one 
molecule changes from time to time, it is assumed that the 
average velocity for a sufficiently large number of molecules remains 
constant. 

These are, briefly stated, the essentials of the kinetic theory of 
gases. Though it is properly called a theory, yet its agreement 
with all available experimental facts is such that it stands unques¬ 
tioned by modem scientists. 


KINETIC THEORY OF GASES 


47 


54. Pressure and Temperature in terms of the Kinetic Theory. 

In order to simplify the reasoning we will make two artificial 
assumptions known to be untrue, namely that the molecules are 
smooth spheres and that the walls upon which they impinge are 
perfectly smooth so that a molecule striking normal to the plane 
of the surface is reflected back upon itself. We may justify these 
assumptions by stating that in the aggregate the effect is the same 
as though the molecules were smooth spheres no matter what 
their shape and considering the aggregate of impacts the effect 
is the same as though the walls were mathematically smooth. 
The fact that results deduced on this basis agree with experi¬ 
mental facts indicates that in the aggregate the effect is the 
same as though these conditions were actual. 

Consider a gas enclosed in a cube the distance along whose 
edge is (s). The molecules are traveling 
hap-hazardly in all directions. We may 
resolve the velocity of each molecule into 
its components along the x, y and z axes. 
If the actual velocity is V\ the components 
are Ui Vi wi and 

v* = u i + v i + w i 

Consider now the component, ui. When 
a molecule of mass, m, strikes the cube 
face it will be reflected with a velocitv, Ui, 
in the reverse direction, the change of 
momentum being 2mui. The molecule 
will travel across the cube and back again 
in the time (2s/ui) and hence the average force exerted on the 
face due to the single molecule is 

m «2 



If we add up the effects on this face for all the molecules we have 

^ m (u? + U 2 +.) 

r = —■ 

s 

F m(u? + u| +.) 

or since p = P. P- ~ 3 

and we may write 

mn (u? + u| +.) 


where n is the total number of molecules in the cube 



















48 


HEAT 


a? + u£ + • • - *. 


u 2 where u 2 is the average square of 


n 

the x-velocity components. 

™ , m n u 2 

Therefore p = -r— 

s 3 

Since the pressure is the same on every face and since we have for 
the other faces 

p = mnv 2 /s 3 , p = mnw 2 /s 3 , it follows that 
u 2 = v 2 = w 2 


Therefore u 2 


= v 2 /3 and since —g- = p (the density) we have 



84 


For mass of gas, M, we have 

pv = -!r- M>j 2 85 

o 

where v here is the volume and v 2 the mean square velocity. 
Combining eq. 85 with the equation 81 we obtain 


3 1 

—MRT = — M^ 2 which indicates the 

proportionality between the average kinetic energy and tempera¬ 
ture. 

55. Avogadro’s Law. Equal volumes of different gases at 
same pressure and temperature contain the same numbers of 
molecules. -*■ 

We may write equation 85 as 

1 2 
pv =—nmr 

o 


where m is the mass of a molecule, n the number of molecules 
in the volume, v. 

For one gas 


PiV, 



vcuvf 

2 


for a second gas 


2 _ mtvl 






KINETIC THEORY OF GASES 
If pressure, volume and temperature are same we have 


49 



56. Brownian Movements. While it is physically impossible 
to see the individual molecules and study their motions, direct 
optical evidence of the existence of such motion is to be found in 
the so-called Brownian movements. If a drop of water containing 
any very finely divided material in suspension such as, for example, 
carmine, is examined under a high power microscope, the individual 
particles will be seen to possess a peculiar jiggling, dancing motion. 
This motion results from the molecular bombardment to which 
they are being subjected. The particles are huge compared with 
the molecules but are caused to jiggle and dance under the innum¬ 
erable impacts. A log floating in the water and struck by rifle 
bullets from all sides would be knocked about in a similar way. 

57. Deviations from Boyles’ Law—Van der Waal’s Equation. 

The kinetic theory applied to actual conditions indicates that 
Boyle’s law must be only an approximation. The molecules 
actually must occupy an appreciable space and it is only the space 
between them which undergoes change with pressure. The 
observed volume, v, therefore is greater than the variable or 
idealized volume by some constant amount, b. Further, the 
idealized theory takes no account of time lost in collision. If one 
molecule is brought to rest and a second takes up the motion, 
time must be required to slow down the one and accelerate the 
other. The number of collisions each molecule makes in its 
passage across the container and back is proportional to the 
density. Likewise the number that are traveling back and forth 
is proportional to the density. Therefore the time lost is pro¬ 
portional to the square of the density. The observed pressure, p, 
therefore is less than the idealized pressure by some quantity 
which is proportional to the square of the density. To get the 

ideal pressure we must add to the observed pressure (ap 2 ) or , 

where a is a constant; p, the density; and, v, the specific 
volume. To get the ideal volume we must subtract from the 

♦This does not follow from our defination of temperature as proportional 
to tne mean kinetic energy for a given gas. It may be shown however that, 
when two gases mix, temperature equlibrium is established when the mean 
kinetic energy of the molecules of one gas equals that of the other, that is, 
on mixing there will be an interchange of energy until this condition is attained. 


Therefore at the same temperature miV i 


Li \ 
2 


nw. 


2 





HEAT 


50 

observed volume the constant, b. The equation of state then 
becomes 


(p + v 2 )( v - b ) = m R T 86 

This is known as Van der Waal’s equation. 

58. The Porous Plug Experiment. Proof of Molecular Attractions. 

The curves of Fig. 39 
indicate the character of 
the isothermals at room 
temperature for various gas¬ 
es on a pv, p diagram. 
Study of these curves shows 
that at high pressures and 
therefore high densities we 
have a straight line with 
a positive slope. Eq. 86 
meets these conditions if b 
is appreciable as compared 

with v, and ^ negligible. 

Under these conditions we have p (v—b) = c or pv =b p + c. 
At low pressures we have in the case of air and CO 2 a straight 
line with a negative slope. Equation 86 meets these conditions if 
the term, b, is negligible, for then we have 

(p + v = c or pv = —k p + c. 

In the so called “porous plug experiment” a gas is allowed to 
expand through a porous plug without doing external work or 
having work done upon it. Under these conditions the product 
of p v before and after the expansion is unchanged. With all 
gases a temperature change would result from such an expansion 
as indicated by Fig. 39. With hydrogen a rise in temperature 
should be obtained, with air and nearlv all other gases a drop. 
The temperature change for a given pressure change, computed on 
the basis of the pv, p curve, is however not obtained by experi¬ 
ment. For air as much as nine times the computed value has been 
found. The observed rise for hydrogen is always much less than the 
computed value. This lack of agreement is taken to indicate that 
work has to be done in the case of an expanding gas against mo¬ 
lecular attractions. The deviation from computed results is, in all 
cases , in such a direction as to indicate attractive forces against 
which work has to be done at the expense of the kinetic energy of 
the gas. 





HEAT AS ENERGY 


51 


59. Mechanical Refrigeration. The cooling of a gas on expan¬ 
sion is made use of in the manufacture of ice, the liquifaction 
of air, and other refrigeration processes. In ice manufacture 
C0 2 or NH 4 is liquified by pressure and the liquid then allowed to 
vaporize under reduced pressure. The work of separating the 
molecules against molecular attraction both fiom the liquid state 
and as vapor is supplied at the expense of the kinetic emergy of the 
molecules and great reduction of temperature occurs. The 
gas after this expansion is compressed again to liquid and the 
process continued. The expansion takes place in coils immersed in 
running brine into which are placed the tanks or containers with 
water to be frozen. The compression occurs in a Cylinder or coil 
which is water-jacketed so that the heat of compression and 
liquifaction is carried awav. As a result of the process heat 
passes into the working substance, C0 2 or NH 4 , from the brine 
in contact with the expansion coils and is carried away by the 
running water about the compression chamber. For refrigeration 
in cold storage plants the chilled brine is circulated thorough pipes 
in the cold storage rooms. 

In the preparation of liquid air the air is not liquified by compres¬ 
sion but is compressed at room temperature to some 2500 lbs. 
per sq. in., the heat generated bv compression removed by cold 
water coils, and the compressed gas then allowed to expand 
rapidlv. The expansion takes place in coils led back through the 
oncoming gas which is continually cooled and finally liquified. 
The heat of compression which was previously removed has to 
be restored as the gas expands and can only be restored at the 
expense of the temperature of the oncoming gas. 

HEAT AS ENERGY 

60. Quantity of Heat. Kinetic theory has shown that tem¬ 
perature is a measure of the average kinetic energy of the mole¬ 
cules. The kinetic energy in the aggregate is what we know as 
the heat possessed by the substance. Quantity of heat therefore 
may properly be expressed in energy units (the erg and the joule, 
etc.). Usually however heat is expressed in calories or B. T. U’s. 

The calorie is the quantity of heat required to raise one gram 
of water one degree centigrade—to be more exact it is the quantity 
required to raise one gram of water from 14^°C to 15p2°C. 

The B. T. U. (British Thermal Unit) is the quantity of heat 
required to raise one pound of water one degree Fahrenheit. One 
B. T. U. equals 252 calories. 

The heat capacity of the body is the quantity of heat required 
to raise the temperature of the body one degree. The “water 
equivalent” is the amount of water which has the same heat ca¬ 
pacity as the body. 


52 


HEAT 


The specific heat of a substance is the heat required to raise the 
temperature of one gram of the substance one degree centigrade— 
or in the English system, one pound of the substance one degree 
Fahrenheit. The specific heat is not a constant for a given sub¬ 
stance but varies somewhat with the temperature. 

61. The First Law of Thermodynamics. We have seen that 
heat is a form of energy and may be expressed in energy units. 
For many years in the early development of the science of physics 
the true nature of heat was not understood. Experiments by 
Joule and others finally showed that there was an equivalence 
between heat expressed in calories and energy consumed or work 
done when the heat was developed mechanically (as by friction). 
These early experiments showed that heat may be converted to 
work or work to heat and that always the same amount of heat is 
produced from a given amount of mechanical work and vice versa. 
The statement of this fact is known as the First Law of Thermo¬ 
dynamics. It is simply an extension of the Law of the Conserva¬ 
tion of Energy to include heat. The number of ergs in a calorie 
(or the number of foot-pounds in a B. T. U.) is called in either 
system of units the mechanical equivalent of heat * 

In mathematical language the First Law of Thermodynamics 
may be written 

dQ = dU + dW 87 

This means that in general when heat, dQ, is added, part is used 
in doing external work, dW, as the substance expands against 
external pressure; and part, dU, in raising the temperature of the 
substance and in some cases doing internal work of separation of 
the molecules against attractive forces. 

PROBLEMS 

1 . Into 12 kg. of water at 30°C. are dropped at the same 
instant 1 kg. of copper at 100°C., 2.5 lbs. of zinc at 180°F. 
and 3.4 lbs. of ice at -10°F. Find the resultant temperature. 
Specific heat of copper, 0.094; of zinc 0.094; of ice 0.51. 

2. What horsepower is required to raise the temperature 
of 100 lbs of water at 40°C to the boiling point in 30 minutes? 


* There are 4.2xl0 7 ergs per calorie or 778 ft. lbs. per B.T.U. 



53 


CHANGE OF STATE 
CHANGE OF STATE 

62. Vaporization. The phenomena of change of state, solid 
to liquid or liquid to vapor, may be readily explained in terms 
of the kinetic theory. In the liquid state the molecules are suppos¬ 
ed to be so close together that their freedom of motion is restricted 
by molecular attractions* The molecules within the mass cannot 
break loose from the attractions of their fellows. They are still 
in rapid motion and those near the surface are continually breaking 
through, forming a vapor in the region above the liquid. These 
molecules constituting the vapor travel hap-hazardly in all di¬ 
rections and as a result a certain proportion must be continually 
re-entering the liquid. We have thus molecules going both ways 
through the surface. At any particular temperature the escape 
of molecules through the surface will go on until a vapor pressure 
is established such that the number returning just balances those 
escaping. This vapor pressure, which we must consider as existing 
within the liquid as well as in the region above, is equal to the differ¬ 
ence between the “bombardment” pressure within the liquid, 
that is, the impact pressure due to the motion of the molecules, 
and the so called “internal” pressure, (molecular attractions).* 
If the temperature of the liquid is raised, the average molecular 
velocity within the liquid increases and more molecules escape 
through the surface. This goes on until the vapor pressure 
above the liquid again balances that within. A vapor in contact 
with its liquid and in equilibrium with it is called a saturated vapor. 
The pressure of a saturated vapor depends upon the temperature. 
Any attempt to compress a saturated vapor results in condensa¬ 
tion. 

Since in the liquid state the molecules are held within the 
influence of molecular attractions, to change to the vapor state 
work must be done against these molecular forces. Work must 
also be done against the external pressure (atmosphere) as a very 
great increase in volume occurs. The heat required to change one 
gram of a substance from liquid to vapor is called the heat of 
vaporization. This heat (energy) is used partly in the performance 
of internal, partly in performance of external work. Equation 
22 applied to this case becomes 

JQv" = U + / pdv 88 

where, since we are dealing with one gram of the material, Q v 

*When once a bubble is formed it is the bombardment pressure, due to the 
molecules impacts from within which holds back the liquid walls. Molecular 
attraction in the form of surface tension tends to make the bubble collapse. 
The difference between these two is the vapor pressure and this must equal 
atmospheric pressure plus the hydrostatic pressure due to the column of liquid 
above the bubble if the bubble is to persist. 



54 


HEAT 


is the heat of vaporization, J the mechanical equivalent of heat, 
U the internal work, / pdv the work against external pressure. 

When the temperature is such that the vapor pressure equals 
the external pressure we are at the “boiling point” of the liquid. 
Any input of heat then produces change of state without rise in 
temperature until the change is entirely effected. It will be 
observed that the boiling point (temperature) is a function of the 
pressure. The pressure above a liquid is usually that due to the 
atmosphere. Hence usually we mean by “boiling point” the 
temperature at which the vapor pressure equals the atmospheric 
pressure. 

The pressure of a saturated water vapor as a function of the 

temperature is indicated by 
line OA (Fig. 40). This 
curve may also be interpreted 
as showing the boiling point 
for various external pressures 
and indicates that the boiling 
point may actually be lowered 
to 0°C by simply reducing the 
external pressure. 

63. Fusion. In the solid 
state we have a still further 
restriction of the motions of the 
molecules. The crystalline 
structure usually characteristic of this state indicates a definite 
arrangement of molecules in positions of equilibrium under very 
powerful molecular forces. The molecules are still in motion, 
probably largely vibration about equilibrium positions. The 
solid state may be entered into only when the “bombardment” 
pressure (temperature) becomes equal to or less than “internal” 
pressure (molecular attractions) plus external pressure. This 

point is the freezing point and depends upon the nature of the 

material and upon the external pressure. 

When heat is added to a solid the temperature rises until the 
melting point is reached after which the further addition of heat 
causes change of state (fusion) without rise in temperature until 
the change is entirely effected. Equatioh 87 in this case becomes 

JQp = U + /pdv 89 

where (since we are dealing with one gram) Qf is the heat of 
fusion; J is the mechanical equivalent of heat; U, the internal 

work required to tear apart the crystals; /pdv, the external work 

due to the change in volume. / pdv may be positive or negative 





CHANGE OF STATE 


55 


depending upon whether fusion entails increase or decrease in 
volume. It is very small as compared with U. 

Even in the case of a solid there still exists a vapor pressure, 
a difference between the “bombardment” pressure and internal 
pressure (attraction). This probably means that there occurs 
an interchange of molecules from group to group, one molecule 
replacing another without disturbing the grouping and hence 
molecules escaping through the surfaces and re-entering as in the 
case of a liquid. There exists a definite equilibrium pressure 
between the vapor and the solid for each temperature. The vapor 
pressure curve for solid and vapor in the case of ice is shown in 
Fig. 40 line OC. 

The line OB indicates pressures at which ice and water may 
exist in equilibrium for various temperatures. To the left of BOC 
we have ice, to the right of COA vapor, between OB and OA water. 

Fig. 41 shows 
the isothermals 
through change of 
state for such a 
substance as water 
or carbon dioxide. 
Above the point 
(C), called the crit¬ 
ical point , there 
is no transition 
from vapor to liq¬ 
uid. Below that 
point there occurs 
on all isothermals 
y a change of volume 

— at constant pres¬ 

sure indicating con¬ 
densation. 


Fig. 42. shows the re¬ 
lation between specific y 
volume and temperature 
for a saturated vapor 
and its liquid. The 
meeting point of the two 
curves is the critical point. 
All distinction between 
vapor and liquid there dis¬ 
appears. This is the same 
as the point (C) on the 
p,v diagram, Fig. 41. 




Fig. 42 









56 


HEAT 


PROBLEMS 

1. If 100 grams of lead cool from 340°C to 327°C in 2 minutes 
and then the temperature remains steady for 25.8 minutes while 
the mass is solidifying, find the heat of fusion of lead assuming 
that heat is lost at a uniform rate and that the specific heat of lead 
at 330°C is .032 cal. per gram. 

2 . Calculate the work done against external pressure when one 
gram of water at 100°C. is changed into steam. Volume of one 
gram of steam (saturated) at 100°C. = 1650 c.c. Taking the 
latent heat of vaporization as 542 calories per gram, compute how 
much of this goes into external work and how much goes into 
internal energy. Ans. 39.2 cal., ext. work. 

3. (a) A jar containing only water and its vapor is sealed up 
and put into a kettle of water which is brought to the boiling point. 
What will the pressure within it become ? 

(b) If the above jar is half full of water and half full of air 
at pressure of 76 cm. and sealed at 20°C and then put into the 
kettle of water which is brought to the boiling point what will the 
pressure within become, neglecting the expansion of the water and 
the glass. 

4. Two liquids A and B are introduced into two barometer 
tubes, the temperature of each being the same. It is noticed 

(1) that in both cases a little of the liquid does not evaporate, 

(2) that the mercury in the tube containing A is more depressed 
than that in the tube containing B. Which liquid would you ex¬ 
pect to have the higher boiling point? Why? 

PROPERTIES OF GASES 

64. Application of the First Law to Gases. In case we supply 
heat to a gas equation 87 becomes 

pdv , dU 

dQ = mSydT + —y— H—j— 90 

J is the mechanical equivalent of heat; S v is the specific 
heat of the gas at constant volume, that is the heat required to 
raise one gram of the gas one degree centigrade without doing 
external work; m is the mass of gas; dT the temperature change; 
dU, the work against internal forces. 

65. The Specific Heats of a Gas. The amount of heat required 
to raise one gram of gas one degree is not a definite thing but 
depends upon the heat required to take care of the work of expan¬ 
sion. We may have an infinite number of specific heats ranging 
in value between 0 and oo. Two of these are of especial import¬ 
ance, the specific heat at constant volume, S v , and the specific 
heat at constant pressure, S p . 




PROPERTIES OF GASES 


57 


If heat is added to a gas and the volume maintained constant, 
the energy is entirely used in raising the temperature of the gas; 
if on the other hand the pressure is not allowed to change, the gas 
must expand and some heat be converted to external work. The 
difference between the two specific heats therefore is equal to the 
work done per gram in producing the expansion against external 
pressure plus the internal work. Expressed in symbols for (m) 
grams of gas through a temperature rise AT, 



91 


PROBLEMS 

1. Show that for a perfect gas 

S p — S v = R/J 


92 



2. The curve shows the 
way water expands on heat¬ 
ing. Which is the greater 
between 0°C. and 4°C., (S p ) 
or (S v ) ? Which is the 
.orrfiAter between 4°C. and 


J ^ sr iff. -- \ zs ° c . greater between 4°C. and 



Fig. 43 


66. Isothermal Changes. If the pressure on the gas in the 


cylinder is changed, say by 
changing the weights (w), Fig. 
44, the volume will change ac¬ 
cordingly. If the changes are 
made by infinitesimal steps the 
flow of heat into or out of the 
cylinder (assuming conducting 
walls) will keep the gas at the 
temperature of its surround¬ 
ings. Such a change is an 
isothermal change. The pro¬ 
cess may be indicated by a 
curve on a pv diagram. The 
line A B gives all possible 
values of p and v for a given 



P 


Fig. 44. 


temperature. If the pv co-ordinates are given by the point (c), 
increasing the pressure will move us up to (d). If the pressure is 
suddenly increased we will leave the line, the gas experiencing 
a rise in temperature. The gas will then more or less slowly 
settle down to equilibrium at (d). It is only when we go slowly 
enough to allow the heat to flow in or out and thus equalize 
the temperature that we keep approximately on the line. The 
line is the locus of all equilibrium states at the given temperature. 
















58 


HEAT 


It is called an isothermal curve. The equation of this curve is 
pv = c, where c is a constant. 

67. Adiabatic Changes. If the cylinder shown above is sur¬ 
rounded by a vacuum or by some good heat insulating material so 
that no heat can flow in or out an increase in pressure will then pro¬ 
duce a rise in temperature 
and vice versa. Such a 
change in which no heat 
is allowed to flow in or out 
is called an adiabatic 
change. The PV curve for 
such a change will be steep¬ 
er than in the case above. 
If the coordinates of pqint 
(c) represents the initial 
conditions of the gas an 
increase in pressure will 
cause the gas eventually to 
settle into equilibrium at 
some point (d) above the 
isothermal A B, since the temperature has risen. Thus 
the adiabatic curve is steeper than the isothermal. The 
equation for an adiabatic change for a perfect gas may be 
obtained readily from the expression for the First Law of Thermo¬ 
dynamics, Eq. 87. Since in this case dQ = 0, 87 becomes 



S T dt + 2.^1 = 0 


93 


Referring to eq 81 and taking differentials on both sides we 
have 

pdv + vdp = RdT 

from which dT = P^ v + y dp 

K 

by 92 R = J(S P — S v ), hence dT = 


pdv + vdp 
J(S P -S v ) 


Substituting this in 93 we have 

pdv + vdp pdv 

j (s p —s v ) + js7 =0 

This may be written in the form 

0 

V V 

By integration we have 


dp Spdv 
P + S v v 


log p + 7 log v = const 







THERMODYNAMICS 


59 


where y = “ • We may write this in the form 
o v 

p V 7 = const 94 

PROBLEMS 

1. A certain quantity of gas initially at 0°C is suddenly com¬ 
pressed to half its volume. What temperature will the gas 
attain? The two specific heats of this gas are .24 and .17 cal. 
per gram. 

2. How much work is obtained in the sudden expansion of 5 
grams of air from 2 atomspheres pressure to one atmosphere 
pressure. The air being initially at 0°C. The density of air at 
0°C is .00129. The ratio of the specific heat of air at constant 
pressure to that at constant volume is 1.403. 

THERMODYNAMICS 

68. The Thermodynamic or Kelvin Work Scale of Tempera¬ 
tures. It is the purpose of the following discussion to develop 
a temperature scale, the Thermodynamic or Work Scale, which 
does not depend upon the behavior of any particular physical 
property of a substance and which is in that sense, a truly absolute 
temperature scale. In scientific work all temperatures must 
ultimately be expressed on this scale. 

Cylic Processes; Reversible and Irreversible Processes. Any 
process in which the working'substance (such as a gas) is carried 
through various changes and finally brought back to its original 
condition is a cyclic process. If the working substance is at all 
times maintained in a state of equilibrium, that is, if the changes 
are not rapid and abrupt thereby producing temperature varia¬ 
tions and turbulent conditions, the process will be at all times 
reversible, that is, reversing the conditions will reverse the process. 
For example, if we have a gas contained in a perfectly conducting 
cylinder and subject it to a pressure which is decreasing constant¬ 
ly by infinitesimal steps an expansion at constant temperature 
ensues. If now we reverse the conditions, that is, continually 
increase the pressure by infinitesimal steps, a contraction at 
constant temperature ensues. This process is indicated by an 
isothermal curve on the pv diagram. Note that reversible 
processes can only be approximately attained. The slower the 
process the more nearly does the reverse path coincide with the 
direct one. Of course we cannot change an equilibrium state 
without disturbing it and thereby departing from the equilibrium 


60 


HEAT 



curve. For instance we can not 
move on the isothermal from 
(a) to (c). We may approxi¬ 
mately keep to the isothermal 
by taking a succession of in¬ 
finitesimal steps as from (a tob), 
waiting each time for the heat 
generated to flow out and the 
temperature to drop back to 
its original value. Only by 
making the steps infinitesimal 
can the process be called isother¬ 
mal. A sudden large increase in 
pressure will cause us to depart 
widely (adc) from the isothermal although eventually settling to point 
(c), decreasing the pressure in the same way will bring us eventu¬ 
ally back to (a) but by a wide swing below the isothermal (cea). 

A reversible process then may be only approximately attained, 
the more perfectly reversible the process is, the more nearly does 
it follow along an equilibrium curve. 

It should be noted that equilibrium curves are not restricted 
to isothermals and adiabatics, any curve representing a process 
carried out at infinitesimal speed would represent a process in 
which the substance is at all times in equilibrium. For instance 
the pressure of a gas might be maintained constant while the tem¬ 
peratures were slowly 
changed causing the 
gas to expand, or, all 
three variables might 
be allowed to change 
and if the changes 
were slow enough, we 
could retrace the path 
by merely reversing 
the conditions. 

69. The Carnot Cy¬ 
cle. The laws of 
isothermal and adia¬ 
batic changes may 
be expressed as we 
have seen in very 

simple mathematical terms and for this reason a reversible cycle 
following isothermal and adiabatic paths is from a theoretical 
standpoint the most convenient and easiest to study. Such a 
cycle is called a Carnot cycle. A device in which the working sub¬ 
stance is carried through such a cycle is called a Carnot engine. 



Fig. 47 







THERMODYNAMICS 


61 






o'i 


HI 


_ 


mm 


m 


ff 

sP 


11 
11 

J L 


A Carnot engine is only theoretically attainable but its study has 
led tojnany deductions of great practical value to the physicist, en¬ 
gineer and chemist. Figure 47 represents a Carnot cycle on a pv 
diagram. The gas whose state originally is represented by point (a) is 
compressed along the isothermal to (b). During this part of the pro¬ 
cess heat must flow 
out. This is repre¬ 
sented diagramatic- 
ally by Fig. 48 (1) 
which shows a cylin¬ 
der in which gas is 
compressed resting 
on a perfectly con¬ 
ducting base at the 
temperature (T 2 ) .The 
gas is then compress- 


......,,,J 


-ed adiabatically from 
b to c, represented by 
Fig. 48 (2) in which 
the cylinder is placed 
over an insulating 
stand so that no heat 
flows in or out. The 
gas is next allowed to 
expand isothermally from c to d, Fig. 47, represented, Fig. 48 (3), by 
the cylinder placed over the conducting stand at the higher tem¬ 
perature (Ti). Finally we have the adiabatic expansion d to c,* 
represented in Fig. 48 (4) by the cylinder again placed on an 
insulating base so that no heat gets in or out. 


Fig. 48 


70. Efficiency of a Carnot Engine. It has been shown that an 
area under a curve on a pv diagram, expressed byj'pdv, represents 

work done during the process indicated by the curve. From (a) 
to (b) to (c), Fig. 47, work is done on the substance represented 
by the area under a b c. From (c) to (d) to (a) work is done by 
the substance indicated by the area under c d a. . The net work 
obtained during the process is therefore the area (abed). Heat 
(H 2 ) passes out of the substance along ab and heat (Hi) passes in 
along cd. The net energy consumed is thus Hi—H 2 . This must 
represent the heat converted to work or 
Hi—H 2 = Area (abed) 

The efficiency is the ratio of heat converted to work to total 
heat taken in. 


Efficiency = 


Hi-H 2 

H x 


95 













































62 


HEAT 


71. The Second Law of Thermodynamics. The second law, 
like the first, is a formulation based upon experience and may 
be stated in a variety of ways. Thus “Heat cannot of itself (that 
is, without the performance of work by some external agency) 
pass from a cold to a warmer body,” Or “It is impossible to con¬ 
struct an engine which will work in a complete cycle, and produce 
no effect except the performance of work and the further cooling 
of the coldest body in the system.” These statements mean 
the same thing. It should be emphasized that this is simply 
based upon observation. No violation has ever been found. 
So firmly has this law stood every test that it is now taken as the 
criterion of future performance. Nothing which would violate 
this law can work. 

72. Carnot’s Theorem. All Reversible Engines Working 
Between the Same Temperatures Have the Same Efficiency and 

no irreversible engine working between the given temperatures 
can have a greater efficiency than a reversible engine. If this can 
be shown, it follows that efficiency depends only upon the tempera¬ 
tures between which the engine works and that the expression 
for efficiency which we may be able to obtain by means of a Carnot 
engine will hold for any reversible engine and will express the 
maximum efficiency attainable between the two temperatures 
considered. The proposition may be proved in the following way. 

Imagine two reversible engines of the same horse-power each 
working between the same two temperatures and between the 
‘same heat source (supply) and the same exhaust tank (sink). 
The less efficient engine takes more heat from the source and 
therefore passes on to the exhaust tank more heat as waste than 
the other. Imagine these engines geared together, the more 
efficient one running the less efficient one backward. The less 
efficient engine is now taking heat from the exhaust tank and takes 
out more heat than the more efficient engine puts in. The net 
work is zero, the work done bv the more efficient engine being all 
used up (converted to heat) in running the other. Finally the 
less efficient engine passes into the supply more heat than the 
more efficient one takes out. As a net result of this whole process 
we have heat flowing from a cold to a hot region without compen¬ 
sation anywhere. This is contrary to the experience embodied 
In the second law and therefore must be impossible. 

Let us imagine these two engines instead of having the same 
horse-power, being so adjusted that they take in. the same amount 
Of heat from the supply when running forward. The less efficient 
one now converts less heat to work and passes on more as waste 
to the exhaust tank. Now let the less efficient engine be run 
backward as before. More heat is taken from the exhaust tank 
than is supplied to it and the work done by the more efficient 


THERMODYNAMICS 


63 


engine is not all used in running the less efficient engine. The 
reservoir receives the same amount of heat that it gives up. As a 
net result of this process we have a self acting engine getting work 
at the expense of the heat of the coldest part of the system. This 
again violates the second law and hence is impossible. 

Suppose an irreversible engine were more efficient than a reversi¬ 
ble engine. We could then adjust them as above and run the 
reversible engine backward. The result would be as above 
either we have heat flowing without compensation from a cold to 
a hot region or we get work at the expense of the heat of the 
coldest part of the system. Hence no irreversible engine can 
have a greater efficiency than a reversible engine working between 
the same temperatures. 

Nothing is said in the above about the nature of the engine. 
Therefore the efficiency of a reversible engine depends merely 
on the temperatures between which the engine works. That is 
Efficiency = f(Ti, T 2 ) 96 

73. The Thermodynamic Temperature Scale. Since the 
efficiency of a reversible engine depends upon temperatures only, 
that is since 

and since Hi merely 
depends upon the 
size of the engine, it 
follows that the work 
obtained from a given 
engine, (i. e. one tak¬ 
ing in heat Hi at 
temperature Ti) is 
a function of the tem¬ 
perature T 2 only. 
This fact is made use 
of to establish the 
thermodynamic or 
y work scale of tempera - 
— ture —just as the ex- 
Fig. 49 pansion of a gas being 

a function of temperature only is made use of to establish the 
gas scale. In the case we are considering if we vary the oper¬ 
ating temperatures of our Carnot engine, always adjusting them 
however so that we may obtain equal quantities of work we may 
arbitrarily assume that these equal work intervals are a measure 
of or represent equal temperature intervals. Every tempera¬ 
ture scale we have considered is based upon just such an arbitrary 






64 


HEAT 


assumption. If we consider the work obtained by an engine 
operating between the steam and ice points as a measure of 
100 degrees, then the degree is a centigrade degree and any tem¬ 
perature interval could be expressed bv the equation 


0r-0 2 = 


Hr-Kb 

H-Hi 


97 


100 


Where Hi—H 2 is the work obtained by the engine operating 

between any temperatures 0i and 0 2 and 1 is 1/100 of the 

work obtained by the engine operating between the steam and 
ice points. With temperatures so indicated efficiency of any 
engine taking in heat, Hi, at temperature, 0i, is directly propor¬ 
tional to temperature interval. That is, 

Efficiency = A (0 X — 0 2 ) 98 

(The symbol 0 will be used to indicate temperatures on this scale.) 

Now let us imagine a Carnot engine taking in heat, Hi, along 
cd, Fig. 49, (the temperature of boiling water), performing a 
certain amount of work (area c d f e) and passing the unconverted 
balance on to the sink one temperature step lower; this residue is 
taken up by another engine of equal work capacity working from 
this step down to the next one. Imagine we have 100 such 
engines working down through the 100 successive temperature 
intervals. The sink or refrigerator of one engine is the supply or 
reservoir of the next below it and each engine takes out from its 
reservoir exactly the amount of heat passed into it from the engine 
above, converts a certain amount of this to work and passes the 
residue on to the engine below. Now let us imagine we continue 
this process on down past the freezing point until all the heat 
originally taken in has been converted to work. In that case 
efficiency of the system of engines is 100 per cent. 

-—- = A (0i — 0 2 ) becomes then 

tii 

1 = A (0i—0 2 ) 

The point where all the heat has been converted to work must 
be the absolute zero. If there were a lower temperature more 
work could be obtained from the original heat than corresponds to 
its mechanical equivalent, thereby violating the First Law of 
Thermodynamics. Hence, since in this case. 



02 = 0 , 






THERMODYNAMICS 


65 


and equation 30 becomes 

Efficiency = — 5 —— 99 

(If the last step does not yield an amount of work equal to the 
other steps this merely means that the freezing point is not an 
integral number of our arbitrary temperature intervals above the 
absolute zero). 

The value of the thermodynamic temperature scale lies in the 
fact that it is a truly absolute scale in the sense that it does not 
depend upon the behavior of any physical property of a sub¬ 
stance. It is therefore chosen as the standard scale, all others 
being referred ultimately to it. 

74. Entropy. If a substance is carried around a Carnot 
cycle we may write the efficiency equation as 
H x — H 2 _ 0i — 02 

Hx 0i 


which may be expressed in the form 


Hi = H 2 
01 02 


100 


Hi and H 2 represent the heat transfers at temperatures 0i and 0 2 

H H 

respectively. The quantities,-^- and—are called the entropy 


changes at temperatures, 0i and 0 2 . It follows from this equation 

and definition that the total en¬ 
tropy change for a Carnot cycle 
is zero. It then follows that the 
entropy change for any reversi¬ 
ble cycle is zero, for any reversi¬ 
ble process may be broken up 
into small isothermal and adia¬ 
batic steps and any reversible 
cycle considered as the sum of 
IT a large number of infinitesimal 
Carnot cycles, Fig. 50. Along 

each adiabatic step the entropy change is zero but along each iso¬ 
thermal step the entropy change is— -g -, and for the complete cy¬ 
cle considered as made up of infinitesimal Carnot cycles the 
total change is zero or 



dH 


= 0 


101 


This equation is sometimes taken as a mathematical statement of 
the second Law of Thermodynamics. It says that the entropy 








66 


HEAT 


change for a reversible cycle is zero. But as no process can be 
actually reversible, since friction and dissipation of heat occur 
with every process known, it follows that every actual process in 
nature is accompanied by an increase in entropy. 

Every conceivable physical process involves an increase in 
entropy either through heat conduction or the conversion of 
potential or kinetic energy or work to heat. In the conduction 
process the heat, dH, still exists as heat but at a lower temperature. 
There has then been an increase in entropy. In the conversion 
of work to heat there has been brought into existance an amount 


of entropy, 


dH 

0 


which did not exist before. 


No process in nature 


can result in a decrease in entropy, the best that can be done is to 
keep the increase as small as possible by making our processes as 
nearly reversible as possible. 

These continual increases in entropy mean continual decreases 
in the supply of energy available for work and continual increases 
in the energy no longer useful or available for work. It means 
a degradation toward some common temperature level for all the 
energy of the universe. When this common level is reached no 
further change or physical process is possible. We are led 
by this sort of reasoning to the disquieting conclusion that the 
universe is running down like a clock that has been wound up 
and the key lost. How it got wound up no one can say. But 
we do know that every process that has been observed involves 
a running down, that is, an increase in entropy. 

75. The Entropy-Temperature Diagram. If entropy, 0, is 
> plotted as abscissa against tem¬ 

perature as ordinates, relation- 

e ships are much more evident 

than on the corresponding p, v 
diagram. For instance isother¬ 
mals are horizontal straight 
lines, adiabatics are vertical 

_ straight lines, and the Carnot 

cycle is a rectangle. Areas on 
<p this diagram mean work just 
as on the p, v diagram. 

76. Comparison of the Perfect Gas Scale with the Thermo¬ 
dynamic Scale. It may be shown that efficiency of a Carnot 
engine is also given by the equation 


Efficiency = - ^ 102 

±2 

where Ti, T 2 are the temperatures indicated on the perfect gas 
scale. 







THERMODYNAMICS 67 

™s being so it follows that the perfect gas scale is identical 
with the thermodynamic scale. 

. Proof of Equation 102 The efficiency of a Carnot engine 
working on the cycle (1, 2, 3, 4) (Fig. 52) between temperatures 
(indicated on the perfect gas scale), Ti and T 2 , is 




Efficiency 

Hi—H 2 
Hj 

i 




H, = 

pdv 

mRTi 

j dv 

J 

I J 

j 



Vj 


Vj 

h 2 = | 

pdv 

mRT 2 

F'dv 

J 

1 J 

J 



V, Va 




Vj 


Since the cycle is made up of two isothermal and two adiabatics 
we have the equations 

P3V 3 = p4V 4 , p 2 v 2 = piVi, p 3 v 3 7 = p 2 v 2 7 , p 4 v 4 7 = piVi 7 
By algebraic transformations it may readily be shown that 

V4 = Vi 

v 3 v 2 














68 


HEAT 


Hence the above equation becomes 

Efficiency = ^ 103 

li 

77. Deviations of Gas Thermometers from the Thermo¬ 
dynamic Scale. We cannot build a Carnot engine and we have no 
perfect gas. Hence we have no direct method of determining a 
true thermodynamic temperature. We can however determine to 
what extent any given gas fails of being a perfect gas. A perfect 
gas obeys the law pv = RT. It exhibits no molecular attractive 
forces, that is, the internal work of free expansion is zero. For a 
perfect gas the specific heat at constant pressure is independent 
of the temperature. Hence by studying pressure and volume 
relations at various temperatures, by performing the “porous 
plug” experiment, to measure the internal work of separation 
when a gas is allowed to expand, and by measuring the specific 
heat at constant pressure under various conditions we may de¬ 
termine to what extent a gas differs from a perfect gas. (For 
further particulars one should consult “Poynting and Thomson’s 
Heat” and other special treatises). 

The actual corrections for the hydrogen and nitrogen constant 
volume gas thermometers (in instruments in which P G = 100 mm. 
Hg.) required for reduction to the thermodynamic scale are 
given below. The deviations of the hydrogen scale for practically 
all purposes are negligible. 


Temperature 

(Centigrade) 

Nitrogen 

Hydrogen 

— 200 

+0.62 

+0.06 

— 100 

+0.10 

+0.010 

+ 10 

—0.002 

— 0.000 

+ 70 

—0.004 

—0.001 

+ 200 

+0.04 

+0.004 

+ 450 

+0.19 

+0.02 

+ 1000 

+0.70 

+0.07 

+ 1200 

+ 1.00 



78. The Use of Thermodynamic Equations. The method of 
reasoning in which a working substance is carried through a Carnot 
cycle and in which the relations established for such a cycle are 
made use of, that is, thermodynamic, reasoning, has led to the 




THERMODYNAMICS 


69 



establishment of many 
relations of theoreti - 
cal and practical im¬ 
portance. As an ex¬ 
ample of this method 
suppose it is desired to 
find the rate of change 
with temperature of 
the pressure of a satu¬ 
rated vapor, having 
given the latent heat 
of vaporization, L, the 
absolute temperature, 
0, and the volume, 
v 2 , of one gram of the 
liquid at the same temperature. Let us imagine our saturated 
vapor to be the working substance of a Carnot engine, and let us 
carry it through the cycle (abed) (Fig. 53). 

The vapor is condensed at temperature, T 2 , heat flowing out as 
we change state indicated on the isothermal from a to b. An 
adiabatic increase in pressure at point b where all is liquid will 
raise the temperature to c. Then an intake of heat at tempera¬ 
ture, Ti, will vaporize the substance, the temperature remaining the 
same until all is again vapor. An adiabatic decrease in pressure 
moves us along an adiabatic curve to a. The cycle is thus com¬ 
pleted and is observed to be a Carnot cycle. 

The efficiency = area (abed) = 

Hi Hi 0i 

where d0 = 0i—0 2 , area (abed) = dp (v 2 — Vi), and Hi = L. 

Hence M 104 


Or d P - L 
d0 (vr-Vi)0 


105 


L must be expressed in work units. If v 2 > Vi as in the case 
of water vapor and water, then d0 is positive when dp is positive, 
or an increase in pressure raises the boiling point. If v 2 cv x , as 
in the case of water and ice, then d0 is negative when dp is positive, 
or an increase in pressure lowers the freezing point. 

We may note the following numerical applications of the above 
as typical of the thermodynamic method 

I. Suppose one gram of steam whose state is represented by 
point a, Fig. 53, is carried through the cycle abed and that the 
temperature difference between the isothermals is 0.1°. We 
have then, since 1 gram steam at 760 mm. has a volume of ap- 











70 


HEAT 


proximately 1651 cc., the following data 

Vapor pressure at 100° = 760 mm. of mercury. 

Vapor pressure at 100°.1 = 762.73 mm. of mercury, 
dp = 2.73 mm. = .273 cm.; 0 = 373, and d 9 = 0.1° 

dp(v2-vi)0 .273 x 13.6 x 980 x 1650 x 373 
Then L = -rr-——-; ergs 


or L = 530 calories /gram, approximately, a number which 
is in fair agreement with the observed value. 

II. We may calculate what would be the lowering of the 
melting point of ice produced by an increase of one atmosphere. 
Let A 9 be the difference in the temperature of melting ice pro¬ 
duced by a change in pressure of one atmosphere. The tempera¬ 
ture of the refrigerator in our cycle is 0° C,. or 273° on the absolute 
scale, and that of the source 273 + A 9. The heat absorbed is 
80 calories or 3.352 x 10 9 ergs, and the work done is 0.0907p ergs, 
where 0.0907 is the change in volume of 1 gram of water on 
freezing. 

W 01—02 p(V2~Vi) _ A 0 

TT _ — eT OT l f “ e 

, 0.0907p _ M 

lhen 3.352 x 10 9 273+5 


Now A 0 is very small compared to 273, so that we shall not 
produce any appreciable error in omitting the term, A0, in the 
denominator of the right-hand member. Thus 
0.0907 x p x 273 = 3.352 x 10 9 t 
p = 1.25 x 10 8 t. 

If p is one atmosphere, or 1013300 dynes per square centimetre, 


1013300 
1.25 x 10 8 


0.0075.° 


This number agrees with the results of experiment. 

79. Maximum Efficiency of a Heat Engine. It has been 
shown that no engine working between two given temperatures can 
have a greater thermal efficiency than a Carnot engine and that 

this maximum efficiency is given by the expression - 1 n 

0i 

The upper limit of possible efficiency is thus fixed for any heat 
engine. It should be noted that in this expression for efficiency 
temperatures must be expression on the thermodynamic scale. 


PROBLEMS 

1. What is the greatest possible efficiency of an engine working 
between the temperatures 140°C, and 20°C. ? 

2. What is the advantage of superheating steam in the case of 
a steam engine ? of using a condenser ? What possible advantages 
might accrue from using mercury as a working substance in place 
of water? This has actually been tried. 










THE TRANSFER OF HEAT 
THE TRANSFER OF HEAT 


71 


80. Heat Transfer. Heat may be transferred by convection, 
the mass movement of hot material; by conduction through 
the material, a molecular process by which the kinetic energy of 
the more rapidly moving molecules of a hot region is passed on 
to the slower moving molecules through molecular impacts; by 
radiation , a wave motion of the same nature as light, probably 
set up in part by the vibration of the electrons of which the atoms 
are composed, in part by the vibration of the atoms in molecular 
groups. The energy of heat radiation is largely made up of the 
long waves in the infra red. 

The heating of our houses by hot air or hot water is an example 
of convection. Convection on a large scale, as seen in the circu¬ 
lation of the earth’s atmosphere, determines our climatic and weath¬ 
er conditions. The transfer of heat through our window panes 
is an example of conduction, although convection plays a very 
important part in this transfer. The transfer of heat from the 
fire-box to the water in a boiler is a matter of conduction although 
both convection and radiation facilitate the process. The actual 
transfer through the walls of the boiler is pure conduction. The 
transfer of heat from the sun is an example of radiation as is the 
transfer of heat from an open grate. Radiation is not heat any 
more than gravitational or electrical energy is heat. It is trans¬ 
formed into heat when incident upon absorbing material. 

The heat transferred by conduction through a body may be 
determined from the law 


H = 


k (Ti—T 2 ) At 

l 


106 


where (Tj — T 2 ) represents the temperature difference between two 
points a distance, Z, apart; A, the cross-section; and t, the time, 
k is a proportionality constant whose value depends upon the 
material. It is called the thermal conductivity of the material. 


THE RADIATION LAWS AND THE MEASUREMENT 
OF HIGH TEMPERATURES 

81. Newton’s Law of Cooling. There are certain quantitative 
relationships or laws between the energy radiated from hot bodies 
and the temperature of such bodies. These laws are the basis 
of high-temperature thermometry. Our entire knowledge of the 
temperature of incandescent bodies depends upon their validity. 
The easiest formulated radiation law is known as Newton's law 
of cooling. It states that for small temperature differences between 
a body and its surroundings the heat radiated, per sq. cm. of sur- 



72 


HEAT 


face, per second, is proportional to the temperature difference. In 
mathematical form 

H = RA(T — T s ) t 

or^=RA(T-T s ) 107 

at 

where H is the heat given off, A, the radiating area, T, the 
temperature of the body, T s , that of the surroundings, t, the 
time, and R, a constant depending upon the material. 

This expresses the rate at which heat is given off. To express 

dH dT 

the rate of cooling one must note that -j- = sm -j- , where s 


dt 

is the specific heat of material, m, the mass and 
cooling. Hence we may write 

I'T' 

-T s ) 


dt 

dT 

dt 


sm § = w- 


dT = RA 
dt sm 


-T s ) 


the rate of 


108 


This law is an approximation, holding only for small temperature 
differences, not to exceed say 25 degrees. 

82. Kirchhoff’s Law. The ratio between the emissivity and 
the absorbing power is the same for all bodies and equal to the 
emissivity of a perfect absorber or “ black-body ” at the temperature 
in question. The emissivity of a body is the energy radiated per 
sq. cm., per second from the body. The absorbing power is the 
ratio of the energy absorbed to that incident. For a perfect 
absorber this is unity. Most surfaces do not absorb all the 
energy incident, part being reflected and part transmitted, the 
amount transmitted depending upon the transparency of the ma¬ 
terial. Surfaces which are black, such as carbon, lamp-black, etc., 
are very nearly perfect absorbers, hence the name “black-body” 
for a perfect absorber. We may write Kirchhoff’s law as 

109 
the 


— =e 

a x ’ 


where e x is the emissivity of the surface for wave length, X; a 
absorbing power for the same wave-length; e^ } a constant for a 
given temperature. e x is the emissivity of a “black-body,” 

; unity the body is a perfect absorber and e x , 
then equals e x . We may write eq. 109 as 


since when a x i 
the emissivity, 
e 

The 


- X' 


e 

ratio £- 

x 


is called the relative emissivity. It 


is equal to the absorbing power and is very nearly independent of 
the temperature. 


RADIATION LAWS 


73 


83. Uniform Temperature Enclosures. It may be shown as a 

consequence of Kirchhoff’s law 
that the radiation from a uniform 
temperature enclosure is full 
or “black-body” radiation. The 
energy incident from without 
upon a small opening into an 
enclosure will necessarily all 
pass in. The absorbing power 
of the enclosure therefore is 
unity and for temperature equi¬ 
librium the emission from the 
opening must by eq. 109 
be that of a “black-body”. It 
will be instructive to subject 
the radiation in such an enclosure 
to the following tests. Let us 
imagine as the most extreme case that the interior surface is 
highly reflecting. Such a surface in the open would radiate 
only a small fraction of the amount of energy that would come 
from a black surface at the same temperature. Now let us place 
a piece of carbon in the enclosure. When the carbon has come 
to temperature equilibrium with its surroundings it will be in a 
state where it is absorbing all the energy incident upon it and 
radiating an equal amount. In other words the stream of radiant 
energy to the carbon from the surroundings will be exactly equal 
to the energy stream away from the carbon. The carbon becomes 
therefore indistinguishable. Likewise a piece of shiny silver 
placed within the enclosure becomes indistinguishable as soon as it 
attains the surrounding temperature. If these facts were not so, 
temperature changes would be occurring within the enclosure and 
this is contrary to our hypothesis. It is a familiar fact that in a 
furnace chamber in which temperature equilibrium is established, 
no outlines or objects can be distinguished. The same is true of 
the chinks between the glowing coals of a grate fire. The radiation 
therefore which escapes through a small opening is independent of 
the material or surfaces within and is characteristic of the temper¬ 
ature only. This fact is of great importance since the general 
radiation laws can be applied only to radiation which is a function 
of the temperature alone. A uniform temperature enclosure ena¬ 
bles us to secure such radiation. 

84. Stefan’s Law. The radiation per sq. cm. from a “black 
body” is proportional to the fourth power of the absolute temperature. 
That is 

e = (7 T 4 

where a is a constant and e is the emissivity. 



110 


74 


HEAT 


If the temperature of the body is Ti and that of the surroundings 
T 2 the body is receiving energy at the rate e 2 = <r T| while radiating 
at the rate ei = <tT\. The net energy radiated in such a case is 
given by the expression 

E = <r(Tt-'H) 111 

This law is the basis of several temperature measuring instru¬ 
ments used in high temperature work. Among these may be 
mentioned the bolometer and the Fery pyrometer. The bolom¬ 
eter consists of a blackened strip of platinum used as one arm of 
a Wheatstone bridge. The radiation received by the blackened 
strip, raises its temperature, the resulting resistance change 
being measured by the bridge and the incident energy computed. 
In the Fery pyrometer the radiation from a furnace opening is 
focussed by means of a mirror upon a sensitive thermo-junction. 
The resulting thermal e. m. f. being a measure of the energy 
received may be interpreted in terms of furnace temperature. 

85. Wien’s Displacement Law. From the time of the ancients 
the color of a hot body has been used as a rough basis of tempera¬ 
ture estimation. In certain industries today a statement of 
color, such as faint red, cherry, orange, white, etc. is a sufficient 
estimate of temperature. There is a physical law behind this 
which was first formulated by Wien, namely, that the wave¬ 
length of maximum energy (see the energy distribution curves, 
Figure 55) shifts toward the shorter wave-lengths as temperature 
rises, and the product of wave-length of maximum energy and 
temperature is constant. That is 

X m T = Const. 112 

The law does not give us a precise method of measuring tempera¬ 
ture because of the practical difficulty in determining the wave¬ 
length of maximum energy. 

86. Wien’s Energy Distribution Law. This law first form¬ 
ulated by Wien and derived on the basis of thermodynamic 
reasoning gives the relation between the energy of any wave¬ 
length, that is, the energy comprised within any narrow band 
of the spectrum, and the temperature, namely, 

Ca 

E = Cl V 5 e ~ XT 113 

X is the wave-length chosen, Ci and c 2 are constants, e is 
the Naperian log. base. 

This law has been found to hold accurately only for the short 
wave-length of the visible spectrum. A modification proposed 



RADIATION LAWS 


75 



.001 .002 003 .004 .005 

''AVE LENGTH /N MILLIMETERS. 
Fig. 55 


more recently by Planck has been 
found to hold accurately over the 
whole spectrum from the extreme 
infra-red to the ultra violet. 
Written according to Planck the 
law is 

= ci X' 5 e( 1- ** ) 114 

For short wave-length this law 
reduces to 113. 

Equation 113 may be written for 
convenience in the form 115 

logioE = ^+K 2 

Ki and K 2 are con¬ 
stants and T is tem¬ 
perature on the ab¬ 
solute centigrade 
scale. This is the 


fundamental law of Optical Pyrometry. 

87. Optical Pyrometry. There are a number of instruments 
which make use of the above law in the measurement of high 
temperatures by optical methods. All make use directly or 
indirectly of some photometric method of comparing the intensity 
of the radiation from a hot body with that from some source 
whose temperature is known. If the ratio of intensities is known 
and the temperature of the comparison source, it is obvious 
from eq. 115 that the unknown temperature may be determined. 
Since it is only necessary to compare intensities and not match 
them it is possible to determine temperatures many times hotter 
than that of the controllable source. In this way the tempera¬ 
ture of the carbon arc has been determined and that of the sun 
and the fixed stars. 

In the Morse pyrometer, one of the best known of these instru¬ 
ments, the temperature of a furnace chamber or hot body is 
obtained by matching the intensity of the radiation from the 
furnace or hot body with that of an incandescent lamp filament 
placed in the line of sight. The current through the lamp is 
adjusted until the filament disappears against the luminous 
background. The intensities are then the same and if both 
bodies are full radiators the temperatures are the same. The 
lamp filament is usually calibrated by sighting on a “black- 




76 


HEAT 


body” whose temperatures are measured by some other means 
such as resistance thermometer or thermo-junction and noting 
the current that corresponds to various temperatures. Tempera¬ 
tures beyond the range of the calibration may be obtained by 
interposing a rotating sectored disk or an absorbing glass be¬ 
tween the lamp and the body whose temperature is sought. In 
this way the intensity to be measured is cut down a definite 
known amount. If the pyrometer is calibrated against a “black- 
body” it only gives true temperatures when the body sighted is on 
“black.” If sighted upon a body radiating freely in the open 
the pyrometer gives what is called the “black-body temperature” 
that is, the temperature which the radiation corresponds to for 
a black-body. The actual temperature in such a case is always 
higher than that given by the pyrometer and may actually be 
300° or more above the pyrometer indication. If the emissivity 
of the surface is known, true temperature may be computed. 

In the Wanner pyrometer one half of a photometric field is 
illuminated by light of a given wave length from the hot body 
whose temperature is sought, the other half by light of the same 
wave length from a comparison lamp. The two halves are ad¬ 
justed to equal intensity by means of a polarizing device, the 
angle through which the Nicol prism is turned being calibrated 
in terms of temperature. The instrument is a direct application 
of Wien’s law. Various other optical and radiation methods are 
in use. One interested should consult, Burgess—“Measurement 
of High Temperatures”, also various Bulletins of the U. S. Bureau 
of Standards notably Vol. 1, pp. 189-255, 1904-5. 


LIGHT 

THE NATURE OF LIGHT 


88. Light as Wave Motion. Light as we know it in physics is a 
form of radiant energy of the same nature as radiant heat. The 
physiological sensation of light is due to the absorption of this 
energy on the retina of the eye. Light is known to be a wave 
motion and as such it has seemed necessary to postulate the exist¬ 
ence of a carrier or medium through which this wave motion is 
propagated. This medium is called the ether. There are two 
theories concerning the mechanism of wave propagation. The 
older theory, now displaced, assumes the ether to be an extremely 
tenuous, yet highly rigid, solid, through which disturbances are 
transmitted as through ordinary elastic material, the velocity 
depending upon the density and elasticity as in any homogenous, 
isotropic, elastic solid. The second theory, which may not be 
inconsistent with the first, considers light to be electromagnetic 
in its nature. 

We may get some insight into the electromagnetic theory by 
approaching the subject from the point of view of the long waves 
of wireless telegraph v. We know how these waves are produced 
and we know something of the behavior of the alternating electric 
and magnetic fields associated with them. The long waves 
of radio telegraphy are sent out by oscillating electric charges, 
either oscillating across a spark gap or in a coil.* In the case 
of a spark discharge the terminals of the gap become alternately 
positive and negative, which means that the electric field oscillates 
in the plane of the gap. These electric displacements and the 
magnetic fields which accompany them are propagated outward 
in all directions from the source of disturbance. They may 
set up oscillations in a second circuit properly attuned, the radiation 
incident upon this second circuit being absorbed. We believe 
radiant heat and light are of ecactly the same nature except that the 
frequencies are much greater, the oscillators in these cases being 
either vibrating atoms which are electrically charged or else 
vibrating electrons. The atoms and electrons may be set to 
vibrating by heat, the long waves (radiant heat) being given off 
at the lower temperatures, the higher frequencies (light) only 
appearing at the higher temperatures when the body has become 
luminous. Vibrations may be set up by other means than heat, 
as by excitation with ultra-violet light, kathode rays, X-rays, 
or by radioactivity. 

We shall not be further concerned with the mechanism of light 
production and propagation. We shall be interested however in 


*See paragraph 243 


77 



78 


LIGHT 


the passage of light through material media and its reflection from 
material surfaces; in other words, we shall be interested in the 
optical properties of materials, and in the reflection, refraction, 
dispersion and interference of light, the laws of mirrors, prisms and 
lenses, and the construction of optical instruments. 

89. Huyghens’ Principle; Rectilinear Propagation. The wave 
theory of light was proposed by Huyghens about 1678. Huyghens 
believed that the wave was propagated outward from the source 
in all directions and that each point in the advancing wave front 
acted as a new source of secondary disturbance. This is the view 
we hold today and is called Huyghens' principle. The propaga¬ 
tion is analogous to that of a shove through a crowd, each individ¬ 
ual receiving the impulse of the shove and in turn becoming 
a secondary source passing the disturbance on in every direction 
except backward. In the case of light waves the disturbance at 
any particular point is almost entirely due to that which comes 
along the direct line from the primary source. A small screen 
placed in front of a candle flame completely cuts off the disturbance 
from the eye. This fact, known as the rectilinear propagation of 
light , was for a long time considered inconsistent with Huyghens’ 
principle. It remained for Thomas Young in 1802 to show that 
all the disturbances reaching a given point from positions off 
the direct line would be so related in phase as to exactly neutralize 
one another. This superposition of light waves out of phase is 
called interference. A demonstration of the interference of light 
made by Young established the wave theory on a firm basis. 
Young’s experiment consisted simply in viewing a distant light 
source between two narrow slits. Each slit acts as a secondary 
source in accordance with Huyghens’ principle while all other 
secondary sources are cut out. A series of colored bands are seen 
when looking through the slits. If a monochromatic source such as a 
sodium flame is used there will appear alternate bright and dark 
bands. The phenomenon is due to the interference of the waves 
from each slit. At a given point on the screen the waves will 
be just out of phase, the effects cancelling and giving a dark 
band. At adjacent points, to either side, the waves will be in 
phase and the effects will add giving a bright band. In the case 
•of white light the various colors or wavelengths interfere construct¬ 
ively or distructivelyat different points on the screen giving a series 
.of colored bands or spectra. No other explanation than that 
<offered on the basis of the wave theory has been able to account 
for these effects. 

90. Color. For any wave motion we have the fundamental 
relation 


v = n X 


116 


REFLECTION 


79 


where v is the velocity with which the wave advances through 
the medium; n, the number of waves passing a given point per 
second or the frequency of the vibrations producing the waves; X, 
the wave-length. The color of a light is due to the particular 
frequency or combination of frequencies emitted. That the 
sensation of color is due to frequency is evidenced by the fact 
that the color of a beam of light does not change on entering 
another medium such as water, whereas it can be shown that the 
wave-length does change. 


REFLECTION 


91. Reflection at a Plane Surface. The reflection of a wave 
from any surface may be shown graphically by drawing successive 
wave fronts, the wave front from a point source in a homogenous 
medium being viewed as an expanding sphere due to the disturb¬ 
ance travelling outward in all directions with the same velocity. 
When this disturbance reaches a reflecting surface its direction is 
reversed at the point of contact. By Huyghens’ principle, viewing 
each point as a new source of disturbance, a reversed wave is 
started out from every point in the reflecting surface as the advanc¬ 
ing wave strikes it. The reversed 
wave-front will be spherical, if the 
reflecting surface is plane, with its 
center as far behind the surface as 
the source is in front. Consulting 
the diagram, Fig. 56, we see that 
ob = ob', b' representing a point 
the advancing wave-front would have 
reached if the reflecting surface had 
not been interposed, b, the corre¬ 
sponding point on the reflected wave- 
front. ob = ob' because the veloci¬ 
ties of the advancing and reflected 
waves are the same. Therefore the 

surfaces (abc) and (ab'c) have equal radii of curvature and the 
reflected wave appears to come from point, s', as far behind the 
reflecting surface as s is in front of it. According to the prin¬ 
ciple of rectilinear propagation an eye placed at E will receive 
a cone of light coming along the axis, or normal to the wave-front 
and therefore appearing to come from point, s'. The angles of 
incidence and reflection, i and r, that is, the angles between 
the normal to the reflecting surface and the incident and reflected 
rays respectively, are equal, as is apparent from the geometry of 
the figure. 

92. Image in Plane Mirror. If we have a luminous object 
as source we may construct for every point in it an image at the 




80 


LIGHT 



corresponding distance behind the mirror. The light entering the 
eye from a, Fig. 57, appears to come along the axis (Eca') and that 
from b along the axis (E db'), whereas actually the axes of the 
wavefronts are (Ec + c a)and (E d + db). Since it is only the 
disturbance along the axis which counts we may consider the light 
entering the eye from a given point as travelling along a beam or 
ray normal to the wave-front. In construction of images it is 
often convenient to use these rays. To construct an image it is 

merely necessary to 
draw two rays from 
each point in the ob¬ 
ject and extend them 
to their intersection 
point behind the mir¬ 
ror. Thus the image 
of every point on the 
object can be located 
and the composite 
gives the image of 
the object. For ex¬ 
ample, the ray, af, 
Fig. 57, striking the 
mirror normally is reflected back on itself, the light appearing to 
travel along the line (a' f a). The ray, ac, striking the mirror 
at c is reflected along the line, Ec, making angle, i, equal to 
angle, r, the light appearing to travel along the line (a' c E). 
The intersection of these two rays at a' locates the image of a 
at that point. Likewise every other point in the image of ab 
may be found. 

93. Reflection at Spherical Surfaces. In reflection from any 
surface an image is obtained only when the reflected wave-front 
is itself spherical. In the case of a spherical mirror the reflected 

wave-front is approxi¬ 
mately spherical only 
when the diameter of 
the mirror is small 
compared with its 
radius of curvature. 
We may find the 
image of a point on 
an object by drawing 
as in the case of a 
plane mirror two rays 
from the point and 
projecting them back 
to their intersection. 
In drawing the re- 



Fig. 58 











REFLECTION 


81 

fleeted ray it is merely necessary to remember that the angles of 
incidence and reflection are equal. The ray parallel to the 
principal axis and the ray through the center of curvature are 
convenient ones to use. The principal axis of the mirror is the 
normal drawn to the center of the mirror surface. All rays 
parallel to the principal axis pass through a common point called 
the principal focus of the mirror. This point is half way between 
the center of curvature and the mirror surface. The proof of these 
statements may be seen from the geometry of the construction 
and is left as an exercise for the student. 

The relation between the distance, u, of the object from the 
mirror and that, v, of the image is given by the equation, 



u and v are known as the conjugate focal distances and f, the 
focal length of the mirror. To establish this relation, it is best to 
have recourse to the construction of wave-fronts, (Huyghen’s 
construction). A simple proposition in geometry showing the 

relation between sagitta of arc and 
radius of curvature will be found 
very useful in this connection. 

94. The Sagitta Theorem. Con¬ 
sider the arc (abc), Fig. 59, with 
chord ac (= 2y), sagitta, x, and 
radius of curvature, r. r 2 = y 2 + 
(r — x) 2 . From this we get 

y 2 

x = — : -. Now if the diameter 

2r — x 

of the surface, in this case the 
chord of the arc, is small compared 
with the radius of curvature, then we may neglect x in comparison 
with r in the denominator of this expression. And we have the 
relation 


2r 

The only condition under which we get a distinct image from 
a mirror is the condition under which this approximation holds. 

95. Conjugate Focal Relations for Concave Spherical Mirror, 
(abc), Fig. 60, represents a section of a spherical mirror. Imagine 
light coming from an object at o. (a e c) represents a section of 
the wave-front whose radius of curvature as it strikes the mirror at 
a and c is u. A section of the reflected wave-front is indicated 
by (a' b c'). The curvature of the w^ave-front is seen to have been 
increased. The radius of curvature of the reflected wave-front 
is v. The point, o', toward which the wave-front converges 


0u 







82 


LIGHT 


is the image of o, since after passing through o' the light appears 
to come from that point. We now employ the sagitta method as 
follows: 

be = fg, (distances travelled by points on the wave-front in same 
time). 

But be = bf — ef and fg = bg — bf, therefore bf — ef = bg — bf. 
These quantities are all sagitta. bf is the sagitta of the mirror 

surface; ef, the sagitta 
of the oncoming wave- 
front; bg, the sagitta 
of the wave-front after 
reflection. We may 
therefore write this 
expression as 

y2 y2 y2 

27 ~ 2u = 2v 
which becomes 

L + ! = J> 

u v r 
The focal length is 
half the radius of 

curvature, i. e., f = — 



_r 

2r ’ 
118 


hence we may write this equation as 

JL + _L = J. 

U V f 


118 


The position of the image relative to the focus corresponding to any 
position of the objeGt can be found at once from this relation. 

A case which may need interpretation is that in which u 
is less than f, in which case v is negative. The meaning 
is that the image is virtual and behind the mirror. It will be 
left as an exercise for the student to construct the image for 
this and other cases. It is merely necessary to follow the pro¬ 
cedure outlined in paragraph 93. It will be necessary in apply¬ 
ing equation 118 to adopt some convention regarding signs. 
The simplest rule is to regard the case of a concave mirror pro¬ 
ducing a real image (the case treated) as the standard case. 
The signs of u, v and f are considered positive if these quantities 
are on the same side of the mirror as in this standard case , negative 
ij on the other side. 

96. Convex Mirror. Consider the mirror (M a N) and the 
wave-front (S a T) coming from source at o and in contact 
with the mirror at point, a. The portion reflected from a will 
travel back to b while the portions at S and T are travelling 
forward to M and N, respectively, distance, ab, being equal to 






REFLECTION 


83 


distances, MS, or, NT. The reflected wave-front will then 
be indicated by the dotted line (M b N.) The method now is to 
equate cd, which is the same as M S, to ab and then write 
these quantities in terms of sagitta and proceed as in paragraph 95. 
Thus 


ca = aD 


But cd = ca + ad and ab = cb — ac. 

Hence ca + ad = cb — ac or 2 ac = cb — ad. 

These quantities are all sagitta and the equation may be written 
2 y 2 y 2 y 2 2 1 1 

2r ~ 2 v 2u ° r 7 “ “T 119 

the opposite side of the mirror from 


4 V 

Now both r and v are on 



the standard case and hence are negative. Therefore, giving r 
and v their negative signs, the equation becomes 


l-i+I 120 

r v u 

which is identical with the standard case. 

This equation indeed is the general equation applicable to 
all cases if the proper sign is given to each of the quantities involved 
Thus, applying it to the case of a convex mirror, we note that r 
is negative which means that v is also negative or that the image 
is virtual or behind the mirror. 


97. The Parabolic Mirror. A spherical wave starting from the 
focus of a parabolic mirror will after reflection be transformed 
to a plane wave, or, in other words, will become a parallel beam. 
This is the principle of the naval searchlight. The proposition 
may be shown as follows. A section of the unreflected wave 
starting from the focus, o, is indicated by the portion of the circle 
(a b c). The secondary waves reflected from points, d, and, e, 
will have travelled respectively distances, df, and, eg, equal 
respectively to distances, df', and, eg', the distances they would 
have travelled if there had been no reflection. Draw D D' 
as the directrix of the parabola. The parabola is the locus of 





84 


LIGHT 


all points equidistant from the focus, o, and the directrix, D D'. 
Therefore em = oe and dn = od. Also we know that oe + eg' = 

same circle. Now substituting me 


V 


'K 


'He 


od + df', being radii of the 
for oe and eg for eg', 
and nd for od and df 
for df' we get nf = mg. 

Likewise all other points 
on the reflected wave-front 
may be shown to be the 
same distance from D D'. 

Therefore the reflected 
wave-front is plane. Nor¬ 
mals to the reflected wave- 
front will therefore indicate 
beams or rays parallel to 
the axis of the mirror. 

98. Spherical Aberration. 

In the case of a mirror of 
large aperture a careful con¬ 
struction will show that 
parallel rays after reflection 
from different portions of 
the mirror do not pass 
through a common focus. 

The curve, AB, Fig. 63, 
formed by the intersections 
of the rays is known as the 
caustic curve. It is only for the very central portion of the 
mirror that the rays intersect approximately at a point. It 
may be well to note here that if the periphery only is used 
the depth of focus is very slight, that is, the focal length indicated 
by the distance from A to the 
M N plane is very nearly the 
same as that indicated by the 
distance c from the M M plane. 

To get an image using the peri¬ 
pheral portion of the mirror it 
is necessary to focus at very 
exactly this distance. On the 
other hand, when using the central 
portion a clear image is obtained 
anywhere within the range, D B. 

This distance through which a 
clear image can be obtained is 
known as depth of focus. The same phenomenon occurs with 
lenses. In order to bring to an approximate focus objects at 


i 

i 

; 

i / 

'■*! 

i 

i 

i 


/ ^ V— 

l < 

) / 



_„ 



Fig. 62 


























REFRACTION 85 

various distances it is necessary to secure depth of focus by 
stopping down the lens or mirror. 

REFRACTION 

99. Snell’s Law. The velocity of light is known to depend 
upon the medium through which it is travelling. It is retarded 
in all material media, the retardation depending upon the material 
and the wave-length. The 
bending of a ray of light 
in passing from air to water 
is a familiar phenomenon. 

Let us consider a plane 
wave, AB, Fig. 64, strik¬ 
ing obliquely the surface of 
some transparent medium 
M N, and let us assume 
we are dealing with some particular wave-length, X. The 
disturbances set up in the medium, M N, at the point, B, do not 
travel so far in a given time as those originating in the first medium 
at the same instant at A. Suppose the disturbance from B 
advances into the medium as far as C while the disturbance 
from A advances to D. The wave-front in the medium is 
indicated by the line. DC. The ratio of the velocities in the two 
media is given by the ratio AD/BC. That is, vi/v 2 =AD/BC. 
But AD/DB = sin i and BC/DB = sin r, where i and r are 
the angles of incidence and refraction, respectively, that is, the 
angles between the normal to the surface and the incident and 
refracted ray, respectively. Therefore 

s = s 4^i= n 121 

v 2 sin r 

This ratio, n, is known as the index of refraction for the two 
media. If the first medium is a vacuum the ratio gives the index 
of refraction, n, for the particular medium. 

This relation is known as Snell’s law, having been first stated 
by Willebrod Snell in 1621. Snell’s statement did not have to do 
with velocities. He merely stated that the ratio of the sines of 
the angles of incidence and refraction was constant for any two 
media. Huyghens showed that the ratio of the sines was also 
equal to the ratio of the velocities. 

It should be noted that the wave-front in passing into a material 
medium is in effect rotated about an axis which lies in the surface 
and also in the plane of the advancing wave-front, (assuming a 
plane wave-front). This means that the incident and refracted 
ray and the normal to the surface at the point of incidence lie in 
the same plane. 

100. Critical Angle. If light travels obliquely from a more 




LIGHT 


86 


refractive medium, (i. e., one in which light travels more slowly), 
to a less refractive one, the ray is bent away from the normal. 
If the ray is to emerge the angle of refraction cannot exceed 90°. 
There is therefore a critical angle of incidence for the beam within 
the more refractive medium beyond which a ray is totally reflected. 
Since the index of refraction is defined as the ratio of the velocity 
in the less refractive medium to that in the more refractive medium 
and since in this case we are following the ray in the reverse 
direction, i. e., from 
the denser to the 
lighter, we must 
write n = sin r /sin i 
where r is the angle 
of emergence. The 
critical angle may be 
obtained from this 

equation by placing 0 

r = 90°. Thus Fig. 65 




Reflection as well as refraction occurs at the boundry between 
two media. At normal incidence a maximum amount of light 
passes into the second medium, only a very small amount being 
reflected back into the first medium. As the angle of incidence 
increases the amount of light reflected increases until at the 
critical angle all is reflected. The intensity of the refracted 
beam thus diminishes and becomes zero at the critical angle. 

101. Refraction through a Plate with Plane Parallel Faces. 
In the passage of a beam of light through a medium with plane, 
parallel faces we get no final deviation of the beam but merely 
a lateral displacement. On entering a denser medium each 
color component of which the 
beam is composed is bent to¬ 
ward the normal an amount 
depending upon the particular 
wave-length. Upon emerging 
each component is bent away 
from the normal an amount 
which exactly compensates and 
restores the beam to its original 
direction. A narrow beam of 
white light on entering the 
medium is thus dispersed into 
the spectral colors. On emer¬ 
ging these colors are, for a thin 
plate, practically recombined or 


* 








REFRACTION 


87 


synthesized again into white light. There will be a slightly 
different lateral displacement however for each color as shown in 
Fig. 66. If the medium is very thick the color components of the 
original beam will not be sensibly superposed and a parallel 
beam in which the colors are sorted out results. This dispersion> 
as it is called, will depend upon the thickness of the plate and upon 
the angle of incidence. The shorter wave lengths (higher fre¬ 
quencies) meet with relatively more retardation and are bent 
through greater angles in the denser medium than the longer. 
The indices of refraction for crown glass for the various colors 
are given in the table. 


1.5112 red 
1.5160 orange 
1.5211 blue 
1.5251 violet 

One may compute the lateral displacement of the red and violet 
rays through a given thickness for any given angle of in¬ 
cidence. For example, to take an extreme case, a narrow 
white beam, incident upon a plate of crown glass 5 cm. 
thick at an angle of 50°, would emerge as a narrow colored band, 
the separation between the red and violet being about 0.4 mm. 
For a wide beam the superposition of the colors from different 
parts of the beam would produce white light except at the edges 
where colored fringes would appear. 

102. Refraction through a Prism. If the faces of a transparent 
medium are not parallel there will be a resultant deviation and 
dispersion of the light beam. The deviation of a particular color 
will depend upon the wave-length, the angle of incidence, the ma¬ 
terial of the prism, and the angle between the faces. In the case 
of a prism there is a definite relation between the index of refrac¬ 
tion for a given color, the angle of minimum deviation, and the 
angle between the refracting faces. The phenomenon of refrac¬ 
tion and dispersion through a prism is made use of in many 
scientific and technical instruments and is therefore of such 
practical importance as to warrant a rather careful study. 

Let us trace the ray, 
ab, Fig. 67, through 
the prism shown, ii 
is the angle of incid¬ 
ence at the first face, 
ri, the angle of re¬ 
fraction. At the 
second face, r 2 is 
the angle of incidence 
and i 2 ,the angle of 



Fig. 67 




88 


LIGHT 


refraction. The letters are used in this way for the sake of 
symmetry. 8 is the angle of deviation and a the refracting 
angle of the prism. It may be shown (paragraph 103) that, when 
8 is a minumim, angle ii = angle i 2 , and angle ri = angle r*. 
The geometry of the figure will show the following relations: 

ii— ri + i 2 — r 2 = 8 or (ii + i 2 ) — (ri +r 2 ) =8 

Also ri -f r 2 = a 

Therefore ii + i 2 = a + & 

If we have adjusted the incident angle ,ii, so that 8 is a minimum 
then ri = r 2 and ii = i 2 or~- = r and g ~t~ ^ = i. 

Zs Z 


at sm i 
N ow -— = n. 
sm r 


+ 5 


Hence n = 


sm 


122 


103. Proof of Proposition that when the Deviation is a Min¬ 
imum the Incident and Emergent Angles are Equal. 

c . sinii sin i 2 

Since -— = n and -— = n we may write 
smri smr 2 J 


. . /sin ii\ , . . /sin i 2 \ 

ri = sm* 1 — j and r 2 = sm* 1 —— J. 

Substituting these relations in the equation, r x + r 2 = oo, we get 



If we differentiate this equation with respect to ii and place 

= o 

di, U ’ 

( remembering from calculus that—sin^u = — 1 
\ dx V i _ u 2 dx/ ' 


we get 


cos li 


cos i 2 


di 2 

dii 


/ sin 2 ii I sin 2 i 2 

n V ~ 

If we differentiate the equation, ii-{-i 2 =d-|-5, with respect to i, we 
















REFRACTION 


89 


, di 2 , 

gt_ dI7 =_1 ' 


Substituting this in the above we have 


cos li 


cos i 2 


I _sin 2 ii / sin 2 i 2 
\ n 2 \ n 2 


This may be cleared of fractions and written in the form 


m = 


cos 4 ii — cos 4 i 2 


123 


COS 4 li — COS 4 1 2 

We know that n is not equal to unity. The only other condition 
under which this equation can be true is that ii = i 2 which makes 


n =5 — or indeterminate from these data. 

104. The Spectrometer. The spectrometer is an instrument 
for measuring wave-lengths and indices of refraction. It is 
composed of a prism, telescope and collimator tube, Fig. 68. 
The prism is mounted on a table which may be rotated about a 
vertical axis. Telescope and collimator may be independently 
rotated about the same axis. The collimator consists of a tube, 
one end of which contains a narrow vertical slit, s, through 
which is admitted the light to be analysed. The other end of 
this tube contains a lens whose purpose is to send a parallel beam 
through the prism. The slit is therefore placed at the focus of 



this lens. The objective lens of the telescope produces an image 
of the slit on a cross-hair in front of the eye-piece. 

In using the spectrometer the prism table and telescope are 
rotated until the angle of deviation is a minimum. 

This angle is read on a scale and from equation 122 the index 
is computed. If it is desired to measure wave-lengths the prism 
is usually set for minimum deviation for the sodium line and the 
telescope rotated until the cross-hair coincides with the image of 
the slit in some other color whose wave-length is known and the 
angle reading noted. This is repeated for various known wave¬ 
lengths and a calibration curve plotted between wave-length 
and angle readings. Any unknown light may thereafter be ana- 









90 


LIGHT 


lysed by merely setting on the various colored images and noting 
the corresponding angle readings. 

DISPERSION 

105. Dispersion and Dispersive Power. We have defined dis¬ 
persion as the separation of light into its component colors and 
have seen that this occurs when light passes obliquely from one 
medium to another. In the case of a prism of small refracting 
angle, a, we may write equation 122 as 


a 

and hence the deviation as 

8 = a (n — 1 ) 125 

If 8 i and 8 3 be the deviations for two wave-lengths, (83 — 81 ) is 
called the dispersion. Using equation 125 we may write for the 
dispersion 

83—81 = (n 3 — m) a 126 

If 82 is the deviation for a wave-length half way between 

81 and 83, the ratio -^-r —- is called the dispersive power. 

02 

The dispersive power, d, is thus the ratio of the dispersion to the 
mean deviation. Using equation 125 we may write for dispersive 
power 

d= n 1 -n L - 127 

n 2 — 1 

The dispersive power of a given material varies in different 
parts of the spectrum. Likewise the dispersive powers of different 
materials (crown and flint glass, for example) vary. There is 
no simple relation between the dispersions in one part of the 
spectrum for two materials and that in another part. Two 
materials may have the same dispersive powers for a region in 
the red and widely different dispersive powers for a region in the 
blue. Such dispersion is said to be irrational. 

The following table shows dispersion, dispersive power, d, 
and index of refraction, n, for a number of substances. The sub¬ 
scripts, v, and, r, refer to lines in the extreme violet and red ends 
of the spectrum respectively. 



8 —8 
v r 

d 

n (for sodium line) 

Water 

.014 a 

.042 

1.333 

CS 2 

.092 a 

.145 

1.630 

Crown glass 

.023 a 

.043 

1.534 

Flint glass 

.036 a 

.061 

1.587 

Rock Salt 

.031 a 

.557 

1.554 





DISPERSION 


91 


106. Dispersion without Deviation; Direct Vision Spectroscope. 

It is desired to construct a direct vision spectroscope, that is, an 
instrument which will disperse a beam of light into its spectrum 
without producing any deviation of the mean wave-length. 
To accomplish this we use two prisms of different material, one 
inverted relative to the other and adjust the refracting angles, a, 
a', so that the mean deviation is zero, that is, so that the devia¬ 
tion of one for the mean ray is neutralized by that of he other. 
We have then 

a(n m — 1) = (n' m — 1) 128 

If the indices of refraction for the mean 
ray, n m , and ,n' m , are known for the 
two materials then for a given refracting 
angle, a, of one prism, the corresponding 
angle, a', for the other may be computed 
from equation 128. When this relation 
is satisfied it will then be found that the dispersions, a (n v — n r ) 
and a' (n' v — n' r ) are not equal. Thus we have dispersion with¬ 
out deviation. 

107. Deviation without Dispersion. It is desired to construct 
an achromatic prism combination, that is, to produce bending of 
a ray without dispersion. We use as before two prisms of different 
materials, one inverted relative to the other and so adjust the angles 
that the dispersion of one is neutralized by that of the other. 
We have in this case 

a (n v — n r ) = a' (n' v — n' r ) 129 

In this case the deviations a (n m — 1) and ot! (n' m — 1) will not be 
equal. 

PROBLEMS 

1 . If the refracting angle of a crown glass prism is 15°, 
what must be the refracting angle of a heavy flint glass 
prism to be used in combination with it so as to get dispersion 
without deviation? Compute the dispersion obtained. 

2 . What angle must the flint glass prism have for an 
achromatic combination with the 15° crown glass prism above 
mentioned? Compute the angle of deviation obtained with 
the combination. 

SPECTRA 

108. Continuous Spectra. If an incandescent solid is viewed 
through a spectroscope a continuous spectrum will be observed, 
that is, one will see a broad colored band, the colors merging into 
one another from violet through blue, green, yellow, orange to red. 
The rainbow is such a spectrum. The spectroscope really shows 
overlapping images of the slit in the various colors. No matter 
how narrow the slit is, the spectrum of an incandescent solid will 





92 


LIGHT 


be continuous, that is, images in all colors of the visible range will 
be present and will overlap. The spectrum from such a source 
will not be confined to the visible region, the region between the 
violet and red, but will be found to extend beyond the red and 
beyond the violet if suitable means of detection are employed. 
For instance, in the infra-red, instead of using the eye as a receiver, 
the radiation may be allowed to fall upon a blackened strip of 
platinum raising its temperature and thereby increasing its electri¬ 
cal resistance. From the observed change in resistance, the tem¬ 
perature change, and, from this, the energy received per second 
may be computed. Or, the energy may fall upon the blackened 
face of a thermopile. The rise of temperature and the heat 
capacity of the thermopile being known, the incident energy 
may be computed. In the ultraviolet the spectrum may be photo¬ 
graphed for a considerable region and intensities measured in 
terms of plate density. 

109. Wave-Length Units: The Spectral Range. Wave lengths 
are measured in microns, micromillimeters, or Angstrom units. 
A micron, for which the symbol, p., is used is a thousandth of a 
millimeter; a micromillimeter, p.p., is a millionth of a millimeter; 
and an Angstrom unit, (sometimes called a tenth-meter) is a ten- 
millionth of a millimeter. Thus the wave-length of light from a 
sodium flame is .5896 p, or 5896.0 Angstrom units. 

The visible spectrum is comprised between the limits .3800p. 
and .7500 p.. Infra-red waves have been measured as far out as 
314 p,, while the shortest electric waves are in the neighborhood of 
4000 p. or 4 mm. X-rays are in the region 0.0001 p., while the 
gamma rays from radium are of the order 0.00001 p.. 

110. Bright Line Spectra. Spectra from incandescent vapors 
are not continuous as are those of solids but consist of isolated 
images of the slit which do not overlap. This indicates that only 
certain frequencies are present. Every element has its character¬ 
istic spectrum. For instance, the mercury arc spectrum in the 
visible region is composed of six bright lines and a number of 
fainter ones. There is a line in the orange, two yellow ones, 
one green and two violet.; the wave-lengths of these lines are 
respectively .6153 p,, .5791 p,, .5770 p., .5461 p., .4359 p, and .4047 p,. 
There are also lines in the ultra violet and infra-red. Indeed 
the longest infra-red wave-length measured, 314 p., is a mercury 
arc line. Sodium vapor shows two strong lines in the yellow at 
.5896 p. and .5890 p.. These are so close together as to appear 
usually as a single line. So definite are the lines corresponding 
to a given element that we have here a most sensitive means of 
analysis. Minute traces of impurity may be detected by vapori¬ 
zing a material in a suitable flame and observing the spectrum. 


SPECTRA 


93 


Lines corresponding to all the elements present will appear and 
may be identified. 

111. Absorption Spectra. If a beam of light from an incandes¬ 
cent solid passes through some medium, water, alcohol, glass, 
solutions of various sorts, vapors, etc., before entering the spectro¬ 
meter slit, the spectrum observed will no longer be continuous 
but will show dark lines or bands indicating the absorption by the 
medium of certain frequencies. These frequencies are usually 
those which the absorbing material will emit when raised to incan¬ 
descence. For instance light passing through sodium vapor, 
which is at a lower temperature than the emitting source, will lose 
the frequencies corresponding to the sodium lines. The sodium 
atoms resonate to these particular frequencies, absorbing the 
corresponding energy from the beam. * The spectra obtained under 
these circumstances are known as absorption spectra . 

The sun’s spectrum when observed with a spectrometer having 
a very narrow slit and high resolving power is found to be crossed 
by hundreds of dark lines. These are known as Fraunhofer lines. 
They were first obesrved by Wallaston in 1802 and carefully 
studied by Fraunhofer in 1815. This is an absorption spectrum; 
the light from the interior of the sun passing through the rela¬ 
tively cooler vapors in the outer envelope, loses the frequencies 
corresponding to the elements contained in these vapors. A 
study of these lines enables us to identify the elements found in the 
sun. The helium absorption lines were found in the sun’s spec¬ 
trum many years before that element was discovered on the earth. 

112 . The Color of Bodies. Most bodies reflect certain colors 
or portions of the spectrum and absorb the remaining portions. 
This gives the body its color. Such bodies when heated radiate 
most freely those frequencies which it was able to absorb when 
cold. These frequencies are the natural frequencies of the 
vibrators(atoms or electrons) of which the material is com¬ 
posed. The reflected waves are of those frequencies which 
the vibrator cannot resonate to. 

113. Luminescence. Many materials may be caused to glow 
or luminesce without any temperature change whatever. Such 
radiation is thought to be due to electrical dsturbance within 
the atom. This disturbance may be induced by chemical change, 
by the absorption of ultra-violet light, by kathode ray bombard¬ 
ment or by various other means. A luminescence which occurs 
only during excitation is called flourescence. Phosphorescence 
is a luminescence which persists after the excitation has ceased. 
Certain materials such as calcium sulphide after exposure to 
sunlight will glow or phosphosesce for hours in the dark. The 
luminescence of material undergoing radioactive changes is familiar 
to every one. Luminescence appears to arise from changes occur- 


94 


LIGHT 


ing within the atom itself and as such its study seems to offer a 
key to the secrets of atomic structure. 

LENSES 

114. Refraction through Lenses: Types of Lenses. There 

are two general types of lenses, convex and concave. The 
former are thicker at the center than at the edges and produce 
therefore a greater retardation of the portion of a wave-front 
passing through the center than of the portion passing through 
near the edges. The reverse is true of concave lenses. With a 
convex lens the curvature of the wave-front is diminished and 
may be reversed; a plane wave will be rendered converging, i. e., 
all portions will converge toward a common point or focus. Con¬ 
vex lenses are thus said to be converging lenses. Concave lenses 
on the other hand produce less retardation of the central portion 



of an advancing wave-front. A plane wave will be rendered 
diverging. Concave lenses are thus called diverging lens. 

Fig. 69 shows types of lenses. (1), (2), (3) are convex lenses; 
(4), (5), (6) are concave lenses. (1) is double convex, (2) is plano¬ 
convex, (3) is concavo-convex, (4) is double-concave, (5) is convexo- 
concave, (6) is plano-concave. 

115. Path of Ray through a Lens: Construction of Image. 

The path of a ray through a lens may be constructed by the use of 

the relation ^ ^ = n . The incident angle, i, and the index 

of refraction, n, being known, the direction of the ray through the 
material can be determined as can likewise that of the emergent 
ray. The construction of any two rays from a given point will 
determine by their intersection the image of that point. There 
are two rays which for very thin lenses may be drawn without 
troublesome computation and measurement. One is the ray 
parallel to the axis of the lens, the other is the ray through the 
optical center. The axis of a lens is the line passing through the 










LENSES 


95 


centers of curvature of the lens faces; the optical center is the 
point on the axis through which a ray will pass without deviation. 
This point is usually within the lens. All rays parallel to the 
principal axis will pass through a common point called the principal 
focus of the lens. Thus the two rays (MFM') and (MOM'), 
Fig. 70, the first parallel to the axis, and passing through the 
principal focus, F, the second, the undeviated ray, passing 
through the optical center, O, determine the position of the 
image of M by their intersection at M'. Likewise the image of 
N can be located at N' and similarly all other points. This 
method of construction may be applied to any type of thin lens. 



116. Conjugate Focal Relations: Huyghens’ Construction for 
Converging Lens. Let us take as our standard case, the double 
convex lens, Fig. 71. Consider the portion of a spherical wave 
RbT, diverging from source ,S. After passing through the lens 
the curvature of the wave-front will be changed as indicated by 
R'dT'. We will now apply the principle that the distances between 
corresponding points in successive positions of an advancing wave- 
front are travelled in equal times. For this particular case it takes 
the disturbance the same time to travel the distance, bd, through 
the lens as it does to travel the distance R k R' through air. For 
small aperture lenses, (i. e., lenses of small area), angle, 8, is small 
and the path R k R' is sensibly equal to the equivalent air path. 



Fig. 71 
















96 


LIGHT 


We are only interested in small aperture lenses, for only with such can 
distinct images be obtained. Let us now write RR'= v a X and bd = v g t, 
where v a is the veolcity of light in air; v g , its velocity in 
glass; t, the time for the wave to travel these equivalent paths. 
From these relations we have RR'/v a = bd/t> g . Now RR' = 
R k+k R'=ac + ce, ac = ab + be, ce = cd + de and bd = 


be + cd. Making these substitutions we have 
be + cd 


ab + be + ce 


or ab + be + cd + de = n (be + cd). Now ab = 


2 u 


the sagitta of the oncoming wave-front, u, being its radius 

y 2 

of curvature, and y being equal to kc; be = —, the sagitta 

v 2 

of the first lens face, n being its radius of curvature; cd = -— 

2r 2 

the sagitta of the second lens surface, r 2 being its radius of curva- 

y2 

ture; de = the sagitta of the emergent wave-front, v being 

its radius of curvature. Making these substitutions, simplifying 
and rearranging terms, we have 

1 

— + 
u v 


L = (n — 1} (L + L) 

V V r2 / 


130 


u and v are said to be conjugate focal distances and points, S, 
and, I, are conjugate foci. 

If the oncoming wave-front is plane, which means all rays 
parallel to the principal axis and therefore passing* through the 
principal focus, the equation gives 


T = (n 


1 ) 


(ir + i) 


131 


since u = oo, and (for this case) v = f. This equation gives the 
focal length of the lens in terms of the radii of curvature of its 
faces and the index of refraction. Equation 130 may now be 
written 




132 


117. Rule of Signs. These equations may be applied to all 
types of lenses if we observe the following rule of signs. Consider 
all quantities, u, v, r\, r 2 , f, positive if they are on the same 
side of the lens as in the standard case of the double convex lens; 
negative if on the other side. 

118. The Diverging Lens. Equations for a diverging lens 
corresponding to 130, 131 and 132 may be obtained by following 





LENSES 


97 


the methods of paragraph 116. Or we may consider equations 
130, 131, 132, general and apply the rule of signs to the special 
cases. Thus, Fig. 72, ri and r 2 are negative. If we have a 
real source, u, is positive. Substituting these signs in eq. 130 
we get 

Likewise, using the proper signs, eq. 131 become 

L = _(n-l)(i- + l) 134 

Hence, for a double concave lens the equation corresponding to 

eq. 132 is — H-- ——z 135 

u v f 

The negative focal length, f, means that a plane wave-front after 
refraction appears to diverge from a point, F, between the source 
and the lens. We have a virtual image at, F. 



Fig. 72 


119. Exercise. The location of images for various positions 
of the object between infinity and the lens for different types of 
lenses is left as an exercise for the student. 

120. Spherical and Chromatic Aberrations. If the refracted 
wave-front is not truly spherical there will be no common point 
of convergence or focus. If the lens is stopped down to very small 
aperture this trouble is minimized. A stop before a lens will leave 
some residual distortion in one direction, while a stop behind the 
lens will leave distortion in the other direction. A stop before and 
behind or a stop between two lenses will practically correct such 
distortion. Often lens surfaces are ground so as to be no longer 
spherical but so as to produce a spherical wave-front for some 
definite set of conjugate foci. Such lenses are said to be aplanatic. 







98 


LIGHT 


By chromatic aberration is meant the non-superposition of the 
images produced by the different colors of which the light is com¬ 
posed. It is due to the fact that the different wave-lengths have 
different indices of refraction. Equation 131 thus indicates that, 
a lens will have a different focus for different wave -lengths. Since 
n is greater for the blue than for the red, the blue image will be 
found nearer the lens than the red. Chromatic aberration may be 
corrected for by using a combination of lenses of slightly different 
optical density such as flint and crown glass. The discussion of 
this point will be taken up later (para. 122). 

121. Combinations of Lenses. If the distance between two 
lenses is small compared with their focal lengths, the focal length 


of the combination is given by 


f fi f, 


For the first lens 


we have ——h——and for the second — 4- — =-^-. U2 

Ui Vi fi U2 V2 f2 

and Vi are equal, if we may neglect the distance between lenses, 
but are opposite in sign since for the second lens the object is to 
the right of it. Therefore adding these two equations we get 

_L+T = J_+J_ 

Ui V 2 fi f 2 


or 



136 


The lenses are usually in contact as in Fig. 73, in which case the 
above proof is rigorous. 



Fig. 73 


122. Achromatic Combination. Lenses of different 
indices of refraction, such as those of crown and flint 
glass, may have the curvature of their faces so ad¬ 
justed that when used in combination the focal lengths 
for the red and violet waves are the same. 

The focal length of the combination for the violet 
is given, Eq. 136, by 

J_,J_L ,x 

f'v + f'v fy (a) 


and for the red by 


f'r 



(b) 


(The primed symbols refer to one lens, the double primed to the 
other). By Eq. 131 we have for the violet, 


-jr = (n' v — 1) k' where k'= ( 

1 v V r 1 r 2 / 


(c) 



OPTICAL INSTRUMENTS 


90 


and 

J- = (n' T - 1) k" where k"= (d) 

Likewise for the red 

fT- = (n'r — 1) k' and -~-= (n’ r - 1) k" (e), (f) 
The condition for achromatism is f v = f r or that—!— = — 

Iv tr 

Hence from equations (a) and (b) above, the condition becomes 

J_ J_ _ 1_ J_ 

fv' + fv" “ f'r + fr g 

Substituting equivalents from (c), (d), (e), (f), we have as the 
condition for achromatism 

(n' v — 1) k' + (n v "— 1) k"= (n' r — l)k' + (n/— l)k' 
or (n v '— n,') k' = (n r n/) k" 

which may be written 

+?) - - ”•■> 0 -+?) 137 

The indices of refraction being known it is necessary to choDse 
the curvature of the lens faces to meet the conditions of this 
equation. 


OPTICAL INSTRUMENTS 

123. Lens as magnifier. If an object O is placed just in¬ 
side the focus, F, an enlarged virtual image, I, will be produced 
at some distance beyond F. If the eye is placed close to the 
lens the position of the lens will naturally be so adjusted as to throw 

the image back to the 
£ distance of most distinct 

JZ ' - = ^ vision where there is least 

fatigue for the eye. For 
the normal eye this dis¬ 
tance is approximately 25 
cm. By the magnification, 
M, is meant the ratio of 
the size of the image to 
that of the object. That 

is, M = -g. Magnification 

is also defined as the ratio of the angles subtended at the eye (or 
the lens) by the image and object when both are at the distance of 



distinct vision. Thus, M = -, Fig. 74. 

P 


The magnifying power 










LIGHT 


100 

may be expressed in terms of the focal length and the distance of 
distinct vision by the relation 

M = 1 + j- 138 

To prove this, note that M = and a = P « “p Substi¬ 
tuting these we get M = -^. Applying the conj ugate focal relation, 

which for this case is — — - = \ ,since v is here negative, and 
u v f 

substituting d for v, we have u = * ^his used in the above 

gives M = ^ ^ ^ or M = 1 + y . 

124. Simple Compound Microscope. A compound microscope 
may be constructed of two lenses, a short focus objective lens, 
which will produce a real image, I, of an object, and a some¬ 



what longer focus eye-piece lens, which will produce an enlarged 
virtual image, I', of the image, I. The eye-piece acts as a 
simple magnifier of the image produced by the objective. 

Actual high power microscopes differ from this mainly in using 
compound lenses to eliminate aberration. 

125. Astronomical Telescope. For a telescope a comparatively 
long focus objective is used and a short focus eye-piece. In this 
case the image produced by the objective is at the focus of the 









OPTICAL INSTRUMENTS 


101 





objective and approximately at the focus of the eye-piece (really 
just inside). The image due to the objective is smaller than the 
object but this image is greatly magnified by the eye-piece which 
again acts as a simple magnifier. The magnifying power is in this 
case equal to the ratio of the focal lengths of objective and eye¬ 
piece. This may be shown as follows. M = -|. g is taken 

here as the angle subtended by the object at the objective lens 
and since the object is a distant one this is sensibly the same as 
the angle subtended at the eye. 

a = g = j- (see Fig. 76). Therefore 

1 e lo 

M = ^ 139 

le 
























102 


LIGHT 


It will be observed also that the distance between lenses is 
approximately the sum of the two focal lengths and that the im¬ 
age is inverted. 

126. Galileo’s Telescope. In the terrestrial telescope devised 
by Galileo a double concave lens is used for the eye-piece. This is 
placed between the image, I', produced by the objective, and 
the objective lens. As in the preceding cases the image due to the 
objective lens is the object for the eye-piece lens, but here we 
have the peculiar circumstance that the object is a virtual 
one. The object (here the image produced by the first lens) has 
apparently moved on through the lens from the right becoming a 
virtual object on the other side. As the object moves on to the 
right, Fig. 77, the image, I", which now is an enlarged inverted 
image of V or an erect one of the original object, recedes and 
reaches the focus again when I' has reached infinity. The 
distance between"the eye-piece and virtual object, I', is s j adjusted 
that the image, I", is at the distance of distinct vision. I' is at 
the focus of the objective and slightly beyond the focus of the eye¬ 
piece so that the distance between lenses is approximately (fi— f2). 

It will be left for the student to show that in this case also the 



e 


INTERFERENCE; DIFFRACTION 


127. The Zone Plate. We have qualitatively explained the 
rectilinear propagation of light as due to the destructive 
interference of the secondary wavelets from all points except 
those along the axis. Let us consider a plane wave od, Fig. 78, 
advancing toward P. Suppose we draw a circle about o 



T 


Fig. 78 




INTERFERENCE 


103 


in the plane of the wave-front of radius, oa, such that for a given 
point, P, the distance, aP, is one-half wave-length greater than 
the distance, oP, or, d, along the axis. That is, aP = d + y 2 \ 
where X is some chosen wave-length. Let us draw another circle 
larger than the first, the radius, ob, of which is such that bP = 
d + X. Likewise, let us draw successive circles whose distances 

from P increase by half wave¬ 
lengths. If now we make a plate 
with such circles and screen out 
the alternate zones ab, cd, ef, 
etc., and place this “zone plate” in 
the path of a plane wave, the 
intensity of the light at P will be 
much increased. The dis¬ 
turbance from the zone, ab, 
is approximately neutralized by 
the disturbance from the zone 
cb, these being just out of phase 
at P. Screening the alternate 
zones thus increases the intensity 
atP since the disturbances from the remaining zones (1), (3), (5), 
etc., are all in phase. The student can easily show that the areas 
of these zones are equal. 

The intensity or energy content of any simple harmonic dis¬ 
turbance is proportional to the square of the amplitude. The 
effective amplitude of the disturbance at P from each zone 
decreases with the distance and the inclination of its path to the 
axis, oP. The difference in amplitude for the first three or four 
zones is much greater that from zone to zone farther out, since 
the angle of the path increases very rapidly for these first few 
zones. Neutralization of disturbances out of phase only results 
when the disturbances are along the same line. Thus while 
screening off all but the first two zones produces a decided diminu¬ 
tion of intensity at P, complete neutralization does not result. 
The resultant amplitude, a, at P is the algebraic sum of the 
amplitudes from each zone, thus 

a = ai — a 2 + a 3 — a 4 + a 5 . 140 

The amplitude of the disturbance from any one zone is 
less than that from the preceding zone but greater than that from 
the following. We may as a very close approximation say that 
the amplitude of any zone is equal to the average of the ampli¬ 
tudes of the preceding and following zones. We may write the 
equation above as 



ai , /ai +a 3 *\ , /a 3 + a 3 \ /“as +a 7 \ 

a = 2 + (~2 - a2 ) + {~2 - a 4 J + (~2 - a 6 J + 141 






104 


LIGHT 


and place each term in parenthesis equal to zero. Thus the 
resultant disturbance at P when the zone plate is not interposed 
has an amplitude which is half that due to the first zone alone. 


a 


ai 
2 


142 


and the resultant intensity at P is one-fourth the intensity due 
to the first zone alone or 



143 


This means that screening off all but the first zone increases 
the intensity at P four times, the intensity then being Ii. 

On the other hand screening the first zone simply makes the 
amplitude, a 2 , from the second zone the first term of the series 
and gives for a, the amplitude at P, 


a = 


£2 

2 


144 


Since a 2 is very nearly equal to ai, the intensity at P is 
practically unchanged. Screening off several of the central 
zones results in only slight diminution of the light at P. This 
constitutes a striking verification of Huyghens’ principle. It 
may be shown that the interposition of the zone plate increases 
the intensity of the light at P, by N 2 times, where N is the 
number of zones.* Study of the illmnination along the axis 
shows that P is not the only bright spot or focus; to the left 
of P will be a dark region where the disturbances from the differ¬ 
ent zones are just out of phase, farther to the left a bright spot 
where they are again in phase. In fact all along the axis inward 

*The amplitude at P when zone plate is interposed is N/2 times the am¬ 
plitude due to the first zone alone, where N/2 is the number of transparent 
zones, (N being the total number of zones). When zone plate is not inter¬ 
posed, amplitude at P is one-half that due to the first zone alone as shown in 


text, i.e., 


a as ai =2a. 


a is the resultant amplitude at P when plate is 


not interposed, a x the amplitude due to the first zone alone. Therefore 


amplitude with plate interposed is-^L X 2a 


Na, or N times the amplitude 


one would have without the plate. The intensity is therefore, with certain 
restrictions to be noted, increased by N 2 times. The increasing inclination 
of the rays as one proceeds from the central zone throws the disturbances 
farther and farther out of line so that although the phases may be alike the 
disturbances very soon cease to be additive. The increase in intensity by N* 
times can only hold therefore for a few zones. 


INTERFERENCE 


105 


from P there alternate bright and dark regions. We also find 
that the bright regions are colored, red being nearest the plate, 
blue farthest. This the student may explain. 

128. Diffraction. When the central zone or several of the 


central zones of a zone plate are screened there is still illumina¬ 
tion at points along the axis as we have seen above. The light 
appears to bend around the edges of the obstacle. It is actually, 
as explained above, due to the disturbances from the outer 
“Fresnel” zones, as they are called. The bending of light passing 
any sharp edge is essentially the same phenomenon. This bend¬ 
ing is called diffraction. The shadow of a thin needle placed in a 
beam from a linear light source will show near its edges alternate 
bright and dark bands or interference fringes. A distant light 
observed through a narrow slit will show on either side a series of 
spectra alternating with dark regions. 

The diffraction fringes due to a straight edge may be understood 
by consulting figure 80. AB indicates an obstacle interposing 
a sharp edge at B to the path of a wave-front advancing to the 
right toward screen, PP 0 . If we select a point, pi, on the screen 


such that the 
distance to it 
from successive 
Fresnel zones, 
drawn with res- 
pect to P 0 , 
differ by single 
wave-lengths, 
the waves from 
the various 
points in any one 
zone will on the 
average differ by 
a half wave-length and complete neutralization will result at pi. 
If we select a point, p2, where the distances to successive zones 
vary by three-halves of a wave-length, waves from the various 
zones will be on the average one wave-length apart and reinforce¬ 
ment will occur. PP 0 represents merely a section of the screen. 
Actually we will obtain alternate bright and dark lines or fringes. 
The bright bands will be colored and diminish in intensity as we 
move into the shadow. Fringes will also be found in the illumi¬ 
nated region just outside the shadow. 






106 


LIGHT 


A 


129. The Diffraction Grat¬ 
ing. Suppose the dots shown 
between A and B, Figs. 
80 and 81, represent lines 
~ or rulings on a glass plate. 
The rulings are supposed to 
be very fine and very close 
together. Imagine this grat¬ 
ing, as it is called, interposed 
in the path of an advancing 
wave. The clear lines be¬ 
tween the rulings will each 
act as independent sources of 
disturbance sending out second¬ 
ary wavelets as shown, Fig. 
82. We will have the origi¬ 
nal wave-front, o a, all points 
of which are in phase. If we construct a plane, o b, such that the 
perpendicular distances from it to the successive slits vary by single 
wave-lengths, all disturbances in this plane are in phase. This 
plane constitutes a wave-front whose axis is opi. Likewise 


3 


Fig. 80 


Fig. 81 



Fig. 82 















POLARIZATION 


107 


the plane, ac, the distances from which to successive slits vary by 
two wave-lengths constitutes a plane in which the disturbances 
are in phase. This wave travels along the axis (o p 2 ). A screen 
S, placed before the grating will show bright regions where it 
intersects the lines o p 0 , o pi, o p 2 , etc. At points between, 
the disturbances will be out of phase resulting in extinction. 
Tf we have white light coming from a narrow slit parallel to the 
rulings, we will get a white image of this slit at S and a series of 
overlapping colored images in the regions of pi, p 2 , etc. If we 
have some source of selective radiation such as a sodium flame or 
mercury arc we get at pi, p 2 , etc., a series of colored lines, a line 
spectra. 

Wave-lengths may readily be measured by means of a grating. 
Let ab, Fig. 83, be the distance between grating lines, called 
the grating space, s. Since the distances from any given 
wave-front to successive slits differ by some whole number of 


wave-lengths be = nX, and — = sin a. 145 



the first, se¬ 
cond or third, etc., from the central spot, d, the grating space, 
is the reciprocal of the number of lines per cm. We may thus 
compute X. 


POLARIZATION 

130. Double Refraction and Polarization. Electromagnetic 
theory indicates that the electrical vibrations which give rise 
to light waves are transverse to the direction of wave propagation 
and that the corresponding disturbance which travels outward 




108 


LIGHT 


as light is a transverse disturbance and not a longitudinal one. 
A light wave from any source must arise from the vibrations 
of countless numbers of electrons. The wave disturbance must 
be thought of, therefore, asmade up of transverse vibrations in all 
possible directions. The fact that light consists of transverse 
and not longitudinal waves is proven by the peculiar phenomena 
of double refraction and polarization exhibited by certain crystals 
such as calcite, tourmaline and others. Light incident obliquely 
on any face of a calcite crystal, for example, will be separated 
into two rays one of which will follow the ordinary laws of refrac¬ 
tion while the other will have an index of refraction depending 
upon the angle of incidence. The first is called the ordinary ray , 
the second the extraordinary ray. The same effect is observed 
for a crystal of tourmaline, except that the ordinary ray is com¬ 
pletely absorbed unless the crystal is very thin. It is found that 
the direction of the extraordinary ray depends upon the direction 
of the crystallographic axis as well as upon the angle of incidence. 
Even at normal incidence the extraordinary ray is deviated unless 
the normal coincides with an optic axis. Since the refraction of 
the extraordinary ray depends upon the direction of the axis of 
the crystal it follows that the velocity of light must depend upon 
the direction of the path relative to the crystallographic axis. 
This is not surprising in view of the fact that the electrical, thermal 
and elastic properties of doubly refracting crystals are different 
in different directions relative to the crystallographic axis. The 
velocities of the ordinary and extraordinary rays are alike in the 
direction of an optic axis but differ for all other directions. There 
must be some peculiar difference in the disturbances which consti¬ 
tute these two rays to account for the different velocities with 
which they travel through the crystal. 

If a crystal of tourmaline is interposed in the path of the two 
rays refracted through calcite it will be found that for a given 
angle of the tourmaline crystal the ordinary ray from calcite is 
transmitted at maximum intensity while the extraordinary ray is 
completely absorbed. As the crystal is rotated the intensity of 
the ordinary ray is diminished and the extraordinary ray begins 
to be transmitted. At 90° rotation the ordinary ray is extin¬ 
guished and the extraordinary ray transmitted at maximum inten¬ 
sity. There seems to be no possible explanation of this effect 
on the basis of longitudinal disturbances, that is, for vibrations 
in the direction of propagation of the wave. On the basis of 
transverse vibrations a very simple explanation is offered, namely, 
that the ordinary ray is made up of all vibrations or components 
in a given direction while the extraordinary ray comprises all 
vibrations or components at right angles to this direction. The 
transmission through tourmaline of the ordinary ray, and when 


POLARIZATION 


109 


rotated through 90°, of the extraordinary ray is thus explained. 
The light of these rays, is said to be polarized. If the transverse 
vibrations on a cord are required to pass through a narrow trans¬ 
verse slit only those in the direction of the slit will be transmitted. 
The waves are polarized. A doubly-refracting crystal acts like 
a slit for light waves. Along the axis all vibrations, and therefore 
both rays, are transmitted equally well. In other directions the 
displacements are divided into two components which travel with 
different velocities and hence are refracted to different extents. The 
ordinary ray travels with the same velocity in all directions. Its 
vibrations are those which are always at right angles to the optic 
axis. In Fig. 84 the vibrations normal to the plane of the paper 
will be perpendicular to the optic axis no matter whether the 
ray travels along that axis or in some other direction, ab. The 

ray made up of these 
vibrations is the ordi¬ 
nary ray and its velocity 
is the same in all direc¬ 
tions. The vibrations 
in the plane of the paper 
are at right angles to the 
optic axis only when 
propagated along that 
axis. If the ray travels 
along ab, the vibration 
in the plane of the paper, transverse to the ray, will not be at right 
angles to the optic axis. They will be propagated with a different 



Fig. 84 


a 


velocity and will be refracted 
more or less than the vibra¬ 
tions normal to the plane of 
the paper. These vibrations 
therefore will be separated 
out into the extraordinary 
beam. 

131. Polorization by Reflec¬ 
tion; Plane of Polorization. 

Light reflected from a mirror 
is found to be partially polar¬ 
ized. The vibrations in the 
plane of the mirror are re¬ 
flected to a greater extent 
than those which penetrate 
the surface. The plane 
determined by the incident 
ray and the normal to the 

surface is called the plane of polarization. It will be noted that 




110 LIGHT 

the reflected vibrations are at right angles to this plane. The term 
is used in connection with polarized light in general, the plane of 
polarization being the plane normal to the direction of the 
vibrations. 

132. The Nicol Prism; the Polariscope. Calcite crystallizes 
in rhombohedric form, the crystal faces being parallelograms 
with angles equal to 101° 55' and 78° 5'. The opposite solid 
obtuse angles (a) and (b), Fig. 84, are formed within three plane 
obtuse angles. Any line drawn to (a) or (b) making equal angles 
with the boundry edges or any line parallel to such a line is an 
axis of the crystal. A plane drawn through the blunt edges, ac, 
and, db, contains the optic axis. Such a plane is called the 
principal section. If the end faces, (top and bottom, Fig. 84), 
are cut off perpendicular to the principal section and then the 
crystals cut in two perpendicular to the principal section and also 
to the new top and bottom surfaces and cemented together again 
with Canada balsam, we will have a Nicol prism, (a c b d) 
Fig. 86, shows the principal section of the prism indicated in 
Fig. 84, cd is a section of the cemented 
surfaces. A ray, M N, incident on the 
face, ad, is separated on entering the 
calcite into the ordinary ray, NO', 
which is totally reflected from the Canada 
balsam, and the extraordinary ray, 
NO, which is transmitted through 
the Canada balsam and leaves the prism 
at Q parallel to the original direction, 
MN. Such a prism thus transmits a 
ray of pure, polarized light. 

The polariscope consists of two Nicol 
placed end to end, the second being 
mounted on an axis so that its principal 
section may be rotated relative to the 
principal section of the first. When the 
principal section of the second is parallel 
to that of the first, the ray is transmitted 
at maximum intensity. When the second 
prism and its principal plane is rotated 
the ray is resolved into an ordinary and 
extraordinary ray. When the principal 
sections are at right angles to each 
other the ray is totally reflected at the 
Canada balsam surface because it then becomes the extraordinary 
ray for this prism. It may be shown that the intensity of the 
transmitted light is proportional to the square of the cosine of 
the angle between the principal sections of the two Nicols. This 


M 



i 


Fig. 86 




POLARIZATION 


111 


instrument has important applications in photometry and related 
fields. 

133. Rotation of the Plane of Polarization. If the nicols of 
a polariscope are crossed, i.e., set for extinction, and a plate of 
quartz cut perpendicular to the axis is inserted between them, 
the ray will be restored. If the second prism, which is called the 
analyser , is now rotated the light may again be extinguished. 
The quartz plate has rotated the plane of polarization through the 
angle through which it was necessary to turn the analyser. Many 
substances, not only crystals but solutions, have this property of 
rotating the plane of polarization. In the case of solutions the 
angle through which the plane of polarization is rotated is pro¬ 
portional to the concentration, assuming a given thickness and 
wave-length. The various sugars, sucrose, levulose, dextrose, etc., 
rotate the plane of polarization to quite different extents. The 
polariscope therefore affords a method of chemically analysing 
such substances. 

The subject of double refraction and polarization is abstruce 
and, except in barest outline, beyond the scope of these notes. 
One should consult such texts as Edser’s Light, Duff’s General 
Physics or Wood’s Optics. 


ELECTRICITY 


ELECTROSTATICS 


134. Topics for Review. Experimental facts of the electri¬ 
fication; conductors and dielectrics; the goldleaf electroscope; 
the Faraday ice-pail experiments. Distribution of charge on 
conductors, effect of points. Definition of unit charge. Assume 
the negative charges only to be capable of movement and explain 
the various electrostatic effects. 

135. Coulomb’s Law. The force between electric charges 
is found by experiment to depend directly upon the size of the 
charges and to vary inversely with the square of the distance 
between them. The law is known as Coulomb’s law of electric 
force. It is expressed thus 



146 


k is a factor depending upon the medium and is usually taken as 
unity for air; k, is the proportionality constant whose value depends 
upon the units chosen. It is customary to choose the units in such a 
way as to make k unity. Thus unit charge is defined as that 
charge which will repel another similar charge of the same size, 
placed one centimeter away in air, with a force of one dyne, 
From the experimental fact that a charge placed within a 
hollow charged sphere experiences no force we may deduce Cou¬ 
lomb’s law. Assume q' such a charge placed at point, o, and q, 



the charge per square 
centimeter on the sur¬ 
face of the sphere. 
Describe about point, 
o, a cone of revolution 
making a small angle, 
dco, and intercepting 


* ^ sphere in the areas, dA, 


, the surface of the 


Fig. 87 


and, dB, respectively. 
The force by which the 
charge,qdA,on the area 
dA,acts on q'is exactly 
equal and opposite to 
the force by which the 
charge, qdB, on dB 
acts on q'. This follows 
from the experimental 


112 


ELECTROSTATICS 113 

fact that the resultant force on q r is zero. From geometery 

dAcosa dB cosa' , . # 

- 12 — = - 12 — (equal solid angles at o). Since dco is smal 


a = a' , henc 




The force between qdA and q' 


must be proportional to qdA, q', and to some unknown power of 
r, i. e., 

dFx = C(q.dA). q'. rj 

where C is the proportionality constant. Likewise the force 
action on q' due to the charge on area dB is 
dF 2 = C(q.dB). q'. r” 

As stated above, experiment shows that 
dFx = dF 2 

dA r 2 

Therefore dAr? = dBr" and since _= J we have 

12 dB r| 

r | r£ 

—j =1-^ or n = — 2. Hence the force between two charges 
r 2 1*1 

is found to be inversely proportional to the square of distance 
between them. 

136. Electric Field and Electric Induction. The force acting 
on a unit electric charge placed at a given point in an electric 
field gives by definition the value of the electric field at the point. 
If the force per unit charge is one dyne the field is said to be of 
unit intensity. Electric fields ase conventionally indicated by 
ines of force or lines drawn in the direction in which a (+) charge 
would move. The number of lines drawn through a sq. cm. 
normal to the field at a given point represents graphically the 
strength of field at the point. Since the value of an electric field 
depends upon the medium, these lines show an abrupt disconti¬ 
nuity in passing from one medium to another. The conception 
of continuous lines independent of the medium and corresponding 
to the field for a vacuum is a useful mathematical artifice. Such 
lines, called lines of induction , do nor represent the field. They 
are proportional to the field, the relation being expressed by the 
equation 

N = kE 147 

where N is called the normal induction; E, the field, and k, as 
will be shown (para. 142) the same constant as occurs in eq. 146. 

137. Gauss’s Theorem. The total number of lines of in¬ 
duction across any closed surface is 4 ir times the charge enclosed 




114 


ELECTRICITY 



Q^ds 


within the surface. The field or 
electric intensity E at a point 
on the surface and due to charge 


q, at o is E = ^ The in¬ 


duction, N, at the same point is 


L, , cl u \ j±±\s oaruo 

N = kE = q/r 2 . 


The total induction due to q 



= qdco since the solid angle 


Fig. 88 


dco = --— . The total indue - 

r 2 


tion over the whole surface due to q therefore is 

:7r 

Total induction = q dco = 47r q 



148 


Likewise total induction due to some other enclosed charge, q', is 
47 rq' or the total induction due to all enclosed charges is times 
the enclosed charge. 

138. Potential. If an electric charge is moved into an electric 
field from some distant point, say some point outside the field, 
(infinity), work will have to be performed against the electric 
forces brought into play. The work required to bring unit 
positive charge from infinity up to any given point in a field is 
called the electric potential at that point. It is the potential 
energy which a unit (+) charge would possess in that position. 

In the case of a charge concentrated at a point and no con¬ 
ducting bodies in the neighborhood, the surrounding field is 
strictly radial. In such a field the potential at a point distant, 

r, from the charge is-^- (q /r). This is obtained by computing the 

work required to bring unit (+) charge up from infinity to the 
point. The force experienced by this unit charge along its path 
is variable and given by Coulomb’s law. Substituting the 
proper expression for the force and integrating the equation dw = 
Fdr between the proper limits, we get the expression for potential 
given above. Thus work to move a unit charge a distance, dr, 



against the force, -, is dw = 


positive charge is negative but a positive force equal to it must 
be exerted in order that work should be done. The total work 




ELECTROSTATICS 115 

to move unit charge from infinity to a point distant, r, from 
charge, q, is 



00 


The difference of potential between two points, distant, ri, and, r 2 , 
respectively from a charge, -fiq, is 

Potential Difference = ^ (q/ri — q/r 2 ) 150 

The student may show that the work done in canying a charge 
from one point to another in an electirc field is independent A of 
the path. 

In case a conducting body is placed in an electric field, electric 
charge will immediately redistribute itself over the body so that 
the potential at all points will be the same. If, when equilibrium 
is established, the potential were not the same at all points, then 
work would have to be done in moving a charge from one point 
to another. This would necessitate the existence of forces over 
the surface of the conductor. Since for equilibrium such forces 
are everywhere neutralized it follows that the potential at all 
points on a conducting body is the same. 

The potential of a conducting body is, by definition of potential, 
the work required to bring a unit (+) charge from infinity (or 
from a place arbitrarily assumed to be at zero potential) and 
communicate it to the body. 

The following relations between potential, V, and electric field, 
E, will be found useful. From definition of potential we may 
write 


V= 



E dx or dV = —E dx or E = — 


dV 

dx 


151 


139. Capacity. It follows from the fact that electric forces 
between charges vary directly with the size of the charges that 
the electric potential of a body varies directly with the charge 
upon it, i. e., 

Q oc V or Q = CV 152 

where C is the proportionality constant. C is called the 
electrical capacity of the body. Capacity is defined as the ratio 
of charge to potential and is numerically equal to the quantity 
of electricity required to raise the potential to unity. 

The capacity of a conductor depends upon the size and shape 
of the conductor and upon certain other factors. For instance a 
small conducting sphere and a large one placed in contact and 


116 


ELECTRICITY 


charged would have the same potential but obviously not the 
same charge. 


140. Condensers. 



Fig. 89 


The effect of neighboring grounded con¬ 
ductors on the capacity of a conductor may 
be seen from the diagram. Without the 
presence of the grounded plate, A, the force 
opposing the approach of additional (+) 
charge on B would be proportional to the 
charge on B. With the presence of A the 
opposing force is diminished by the at¬ 
traction due to the (—) charge on A. There¬ 
fore the potential (or work required to 
bring additional unit charge to B is lowered 
by the presence of A. Therefore a much 
larger charge is required to bring B to the 
given unit potential or the capacity of plate, 


B, is increased. Such a combination for accumulating or storing 
electric charge is called a condenser.* 


141. Dielectric Constant. Experiment shows that the capaci¬ 
ty of a condenser depends upon the medium between the plates 
(the dielectric). Suppose the plate condenser, Fig. 89, has ca¬ 
pacity, C a , when air is the dielectric. With some other dielectric, 
(paraffin, rubber, sulphur), the capacity, C x , = kC a , where k is a 
multiplying factor, its value depending on the dielectric, k is 
called the dielectric constant, or the specific inductive capacity. 
It will be shown para. 142 that the constant of Coulomb’s law 
(eq. 146) is also the dielectric constant. 

The effect of the insertion of a dielectric either partially or 
completely filling the space between the condenser plates is to 
weaken the attractive forces thereby causing some of the (+) 
charge to move to the back of the plate and a general redistri¬ 
bution of the (—) charge more widely spaced over the ground. 
There results then a wider spacial distribution of the force lines 
and longer paths along which work must be done to get from the 
ground to the back of the positive plate. Since the work to get 
a unit charge across the space between the plates is less, it is less 
by any path and therefore the potential is less which means that 
the capacity has been increased. 


*B is not charged by carrying (+) charges across the condenser space 
from A but is charged by some external path. The work per unit charge 
across the condenser space from A to B would be the same as by any path 
from the ground to B. The electric force is greatly increased as A approach¬ 
es B but the path is shorter and, since by an external path the work is less, 
the work must be less by the internal path or the shortened path must more 
than compensate for the increased force. 






ELECTROSTATICS 


117 


142. Units. If a condenser requires one electrostatic unit of 
charge or quantity to raise its potential by one electrostatic, 
unit of potential, it is said to have a capacity of one electrostatic 
unit. If the practical units of quantity and potential are used, 
namely the coulomb and volt (to be defined later) then the practi¬ 
cal unit of capacity is the number of coulomb’s required per volt 
rise in potential. This unit is called the farad. One millionth 
of this, the micro-farad, is the more convenient and usual unit of 
capacity. 

It may be shown that the dielectric constant, k , and the pro¬ 
portionality constant, k , appearing in Coulomb’ slaw, (equa¬ 
tion 146), are identical. For example, the potential of a charged 
sphere may be given by the expression V = Q /C, also we may 
write it, by definition, as work to bring unit charge from infinity 
to the surface, assuming that the conditions are the same as 
though the charge were concentrated at the center of the sphere. 



But V = hence C = kr 


Now r is the capacity of a sphere with air as the surrounding 
medium (see next paragraph), and kr is the capacity with some 
other medium for which k is the dielectric constant. Thus 
these two constants are identical. 

143. Capacity of a Sphere. If we do not wish to assume the 
charge concentrated at the center as above we may determine the 
capacity of a sphere by use of Gauss’s theorem as follows. Let 
the field at surface of sphere be E. Then induction, N, — kE 
and total induction = 47rr 2 £E. But by Gauss’ theorem total 
induction = 47rQ, hence 


47rr 2 £E = 47rQ 


or E = 


_Q 

kr 2 


Also since V 



we have V 


l Q 

k r ' 


And since C 



by substituting for V we 


have 


C = kr 


153 


118 


ELECTRICITY 


144. Capacity of a Plate Condenser. The field near an infi¬ 
nite conducting sheet charged with surface density, q, is E and 
is normal to the surface, such surface being an equipotential 

surface. Consider the region en¬ 
closed by the prism shown, one end 
of which projects into the metal 
sheet but not through it. The sides 
of this prism are normal to the 
sheet, the ends parallel. There is 
no force within the metal, all points 
in the metal being at the same 
potential. The total induction 
through the prism is 47 r q da, where 
da is the area of the sheet enclosed 
by the prism and q, the charge 
per unit area. This induction must 
all pass through the end which is 
outside of the metal. The flux 
across this end is also expressed by 
k E da, where E is the field. There¬ 
fore we may write 

Fig. 90 4x q da = k E da ^ _ 4x q 

Since V = — | Eds, we have V = 



■/ 



where; s is the distance between plates. And since C = Q/V, 


C = 


ak 

47TS 


154 


145. Capacity of a Spherical Condenser (Concentric Spheres). 

Assume the inner sphere charged, the outer sphere grounded and 
consider an imaginary spherical surface of radius, r, between the 
two concentric condenser plates. The normal induction, N, across 
this surface is 


AT 4xQ Q c . ^ N T 7 Q 

N = -— 9 = -r Since E = -=- we have E = ^ 

4xr 2 r 2 . k kr 2 



Q_ _ Qfa—n) 

feri k TiTz 


where ri is the radius of the inner sphere, r 2 , that of the outer one. 








ELECTROSTATICS 


119 



Therefore C = -2- = 155 

V r 2 — ri 

146. Capacity of a Cylindrical 
Condenser (Submarine Cable). 

Following the same line of reason¬ 
ing as in the case of concentric sphere 
we have 


N 


4x.Q 2Q 2Q 

n - 7 = T and F = — 

2 x r l tI krl 


And 



147. Capacity of Condensers in Parallel and in Series. 

(a) Series arrangement. 

Charge from the cir- 
- - - - cuit is stored on the first 

and last plates only. 

The intermediate plates 
Pig. 92 become charged merely 

by induction. The 
charge on the second condenser is an induced charge equal to that 
on the first, likewise the charge on the third equals that on the 
second, etc. There is a potential drop across each condenser 
equal to Q/C. The total potential difference across the series is 
the sum of the successive potential drops across the individual 
condensers. Thus 

Pd = Pdi + Pd 2 + Pd 3 + • * * 

Q _ Qi I Q? . Q 3 , . .where C is the capacity 
C Ci 'C 2 C 3 of the combination. 


But as stated above Q = Qi = Q2 = Q3, therefore 


C 


= -i + I + • 

Q ^ C 2 ^ C 3 ^ 


157 













120 


ELECTRICITY 



(b) Parallel arrangement. 

This arrangement is merely.the 
equivalent of increasing the area of 
the plates. The total charge here 
is the sum of the charges on the in¬ 
dividual condensers, i. e. 

Q = Qi + Q 2 + Q 3 + • • • 

This may be written 

PdC = PdiCi + Pd 2 C 2 + Pd 3 C 3 + . 

All the positive plates, being me¬ 
tallically connected are at the same 
potential. Likewise, the negative 
plates. Thus each condenser shows 
the same potential difference. That 


FiR. 93 is, Pd = Pdi = Pd 2 = Pd 3 • • • 

Therefore C = Ci + C 2 -J- C 3 +• • • 158 

148. Energy of a Charged Condenser. The energy of a 
charged condenser is E = 1/2 (Q 2 /C). As charge accumulates on 
a condenser the force against which work must be done to increase 
the charge continually increases. -Suppose at a certain time the 
potential has reached the value V. Then to bring up additional 
charge, dq, the works is dw = Vdq 


, • Q , Qdq 

and since V = ^ dw = -^r 2 - 


or w 


fQdq = 1 ( Q 2 

J C 2 ^C> 


159 


MAGNETISM 


149. Topics for Review. The student should review the more 
obvious phenomena of magnetism, such as the general behavior 
of magnets, location and peculiarities of the poles, influence of 
one magnet on another. He should note what happens when a 
magnet is cut in two, and if this process is carried on indefinitely 
to what conclusion we are led concerning the nature of magnetism. 
He should be familiar with figures obtained by iron filings and 
understand what they mean. 

150. Coulomb’s Law. The strength of a magnet pole may be 
determined by the force exerted at a certain distance away upon 
some pole taken as a standard. It is found experimentally that 
the force between two poles is directly proportional to the 
pole strength of each pole and is inversely proportional to the 
square of the distance between them. In mathematical language 
F oc mm' /r 2 or 



mm' 

r 2 


160 


This is known as Coulomb’s law. The proportionality constant, 









MAGNETISM 


121 


k, depends upon the units only. (1 Ip) is a factor which depends 
upon the medium and is taken as unity for a vacuum. 

By means of this law as stated in equation 160 we may obtain 
a definition of unit pole. Thus, if two equal magnetic poles are 
of such strength that the force exerted between them is one dyne 
when they are placed one cm. apart in a vacuum, such poles are 
by agreement taken as unit poles. With such units k becomes 
unity and is usually not expressed. 

151. Magnetic Field. Magnetic Induction. Lines of Force. 
A region in which magnetic forces exist is called a magnetic field. 
The magnetic field at a point is quantitatively defined as the force 
acting on a unit positive pole when placed at the point. Its di¬ 
rection is the direction of that force. If the force is one dyne 
the field is said to be of unit intensity. This unit of field strength 
is called the “gauss.” It is customary to picture “lines of force” 
as radiating from every magnetic pole. These lines indicate the 
intensity and direction of the field, the intensity being indicated 
by the number of such lines constructed through unit area normal 
to the field. Thus, a field intensity of four gausses is represented 
by four lines per sq. cm. normal to the field (Fig. 94). 

Magnetic fields are found not only in the neighborhood of free 
magnetic poles but are also always an accompaniment 
of electric current. A conductor carrying 
current may be shown to be surrounded 
by a magnetic field, the lines of force in 
the case of a long straight wire being con¬ 
centric circles with the conductor at the 
center. In case the medium in which the 
magnet pole or the conductor carrying electric current is immersed 
is itself magnetizable the lines of force across any unit area normal 
to the field consist not only of those which would be there if the 
medium were a vacuum but also of those due to the local magneti- 

of force as extending 
from one magnet¬ 
ized molecule of the 
medium to the next 
and the magnetized 
molecules as ar¬ 
ranged end to end 
and in rows follow¬ 
ing the direction of 
the magnetic force. 
Such a state of 
affairs is indicated 
in Fig. 95, which 
represents the magnetized molecules of the medium as magnets 



a 

Fig. 95 



Fig. 94 










































122 


ELECTRICITY 


arranged end to end under the action of the magnetic force, this 
force being the magnetic field. Let us imagine a test pole placed 
out among these magnetized molecules without producing any 
modification in their alignment. The effect of the local field 
due to the magnetization of the medium would merely be to move 
our test pole to the negative face of the nearest molecule where 
it would come to rest. The effect of the magnetizing field on 
the other hand would be to carry our test pole along a path be¬ 
tween the rows of molecules in the direction of the field. The un¬ 
balanced force therefore would be only that due to the magnet¬ 
izing field. Since we have defined magnetic field as the force per 
unit pole it is convenient to designate the total number of lines by 
another name, “induction ,” B, and reserve the name “field" 
H, for those lines which represent the actual resultant force 
upon a unit pole. Induction , B, is sometimes defined as the 
force acting on unit pole when placed in a narrow transverse gap 
between the magnetized molecules (a a, Fig. 95), while field, H, 
is the force acting on unit pole when placed in a long longitudinal 
hole threading between the molecules (b b, Fig. 95). It will be 
noted that the induction, B, is made up of two sets of lines, those 
due to the magnetizing field and those due to the local magneti¬ 
zation. 

152. Gauss’s Theorem. The total flux of induction through 
any closed surface is 4 ir times the number of free poles within the 
surface. This fact results fiom the inverse square law and the 
definition of a line of induction. We may see for a simple case 
where the medium is air that this is true. The force at a distance, 
r cm., from a pole of m units is m/r 2 . We therefore draw 
through one sq. cm. at this distance away m /r 2 lines. Through 
every sq, cm. at the distance, r, out from m we imagine this 

number of lines. This gives a total of 47rr 2 ( ) lines radiating 

from the pole, m. Or the total flux, 0, is 

0 = 47rm 161 

The following is a more general treatment. 

(a) Proof for non-magnetizable medium . Imagine a surface 
drawn so as to enclose the + poles m at the end of a magnet. 
The total induction from pole, mi, across any element of surface ds, 

making an angle,, 0, with the normal to the field is —cos 0 or 

r* 

midco, since — C ° S equals the solid angle dco. The total induc¬ 


tion through any closed surface due to pole mi is therefore 


0i=mi 


f* 4 tt 

Jr 


47rmi 




MAGNETISM 


123 



Likewise for any other pole, 
m 2 , enclosed within the sur¬ 
face, </> 2 = 47rm 2 . The total 
induction, <f>, is 0i -f 0 2 
+• • • or 

</>= 47rm 

where m = mi + m 2 
+ ■ • • or the sum of all 
the free poles within the 
surface. 

(b) Proof for magnetiz 
able medium. In the case 


of a magnetizable medium certain additional considerations 
must be taken into account. Draw the Gauss surface as 
before except that outside the iron the surface should be drawn 



so as to pass between the mag¬ 
netized molecules of the medium. 
The magnetization of the medi¬ 
um depends upon the magnetiz¬ 


ing force which is 


m r 


Mor 


where 


Fig. 97 


mo stands for the poles of the 
medium exposed on the inter¬ 
face between the medium and 
the magnet end facing the mag¬ 
net pole, mi, and /jl 0 , the con¬ 
stant (permeability) for a 
vacuum. 

We may picture the magnetization of the medium as a partial 
alignment of the molecules radially from the pole, mo. The 
intensity of magnetization in the medium, i. e., the number of 
unit poles per sq, cm. at a distance, r, from m^ we will indicate 
by the symbol, A. As shown above the total flux of induction 
from every unit pole is 47r lines. Hence the flux from the A poles 
is 47 tA lines. It can be shown that the resultant tangential 
component of these lines with reference to an area normal to the 
^eld is zero. Hence the normal induction due to the local mag- 



124 


ELECTRICITY 


netization is here also equal to the total flux from that source. 
The total induction from all sources per sq. cm. at distance, r, 
from the pole face is then made up of three sets of lines, viz., 
those from the magnet pole, mi, those from the adjoining free 
poles of the medium, m 0 , and those due to the local magnetization. 
Therefore we may write 


B = 


47rmi 47rm 0 


47rA. 


162 


47rr 2 47rr 2 

If there are various poles scattered at different points within the 
surface the equation can be generalized thus 


B = 2 


m 


v m 0 


+ 47rA 


163 


where A is the resultant intensity of magnetization, and 



and 


s nu 

r 2 


are the vector sums of the inductions from the several sources. 


By Coulomb’s law 2 — 2 = H. u 0 is the constant for a 

MoT 2 Mor 2 

vacuum, the medium in our hypothetical long narrow gap be¬ 
tween the molecules. Expression 163 may now be written 
B = ju 0 H -f- 47rA 164 

We will now show that whenever the magnetizing field is due 

to free magnet poles this expression reduces to the form B = 2 

or that the terms which have to do with the medium cancel out. 
At a given pole face, mi, the magnetization of the medium due 
to this pole is mo. The intensity of magnetization, A, at a 

distance, r, from the pole face is since the m 0 poles are at 

this distance distributed over an area 47rr 2 . The field at this 


distance is — -~ r Substituting these expressions for A 

and H in 164 we have B = The same reasoning applies 
to every other pole within the surface, hence 


B = 



165 


The terms referring to the medium have cancelled out and it 
thus appears that B depends only upon the number of free 
poles with the surface and their distribution and is in no way 
effected by the medium. In case the magnetizing field is due 
to electric current in a wire eq. 164 does not reduce to 165 
and it follows that the induction does in that case depend upon 
the medium (para. 154). 




MAGNETISM 


125 

If now we follow the method of part (a) above, we find that the 
total flux of induction in the case of a magnetizable medium is 
also </> = 47rm, where m represents as before all the free poles 
within the surface. 

154. Applications of Equations 162 and 163 to special 

cases. Case of a magnet¬ 
ized iron ring—Induction 
and field in a transverse 
gap. If a magnetizing coil 
is wound uniformly on the 
3r ring as shown, the magnet- 
izing field, H, is merely 
that produced by the current 
in the coil. There are no 
free poles in the magnet to 
produce any modification 
in this field, i. e., the 
second term on the right 
Fig. 98 in eq. 163 is zero. The 

induction is greater than H by the term 47rA. If the current is 
reduced to zero, H becomes zero while B = 47rA. 

Case of a straight iron bar. Here the second term on the 
right of eq. 163 is not zero. All three terms are present. The 
magnetizing field cannot readily be computed since it depends 
upon the magnetization of the iron. It should be carefully 
noted that H, the magnetizing field, is always the resultant of 
the field due to the coil and that due to the free poles of the iron. 

Case of bunched windings on an iron ring. Magnetic Leakage. 

The actual field due to the 
bunched winding is indicated 
by the dotted lines. Only 
partial alignment of the mole¬ 
cules around the iron path is 
effected and free poles are 
produced on the surface of 
the ring. The resultant 
magnetizing field cannot be 
readily computed. It is som- 
what less than in the case of 
uniform winding. 

Case of field due to current 
in a wire. Even though we 
have a magnetizable medium surrounding the wire there are no 
free poles and there is no discontinuity of the medium corre¬ 
sponding to that at the pole face of a magnet. Hence the magnet¬ 
izing field, H, at any point is not modified and may be computed 



Fig. 99 









126 


ELECTRICITY 


from the current in the wire. The circular iron ring with uniform 
winding is a special case of this. For a non-magnetizable medium 
the intensity of magnetization is zero and B = H (from eq. 
163). 

155. Force on a wire carrying current. The general 
case is that of a wire in a magnetizable medium and an example 
is that of a wire in a transverse gap in a magnetized iron ring. 
The reaction is here between the field due to the current and the 
entire local field which is (H + 47rA) lines. 

156. Permeability. In the case of a pole, m, concentrated 


at a point, the field is given by Coulomb’s law as H = 

a number of poles H = — 2-^- 
/x r 2 


or for 


By equation 165 we have B = 2 — Hence 

B = /xH 166 

H is called the permeability of the medium. 

Substituting for B in eq. 164 we get an equation which is 
sometimes useful, namely, 

M = Mo + • -ft is called the susceptibility, k. Hence 


fx= /x o + 47rk 167 

157. Torque experienced by a Magnet. When a magnet is 
placed in a magnetic field it tends to set itself with its axis in the 

direction of the field. Why? 
If the field is uniform it merely 
experiences a turning moment 
or couple. The student should 
indicate the direction and magni¬ 
tude of the forces acting on each 
end of the magnet and show that 
the value of the couple is 



L = mH l sin 6 168 

where l sin 9 is the arm of the couple, l being the length of 
the magnet (i. e., between the poles). If a magnet is placed at 
right angles to a field of unit strength, expression 168 becomes 

L = ml 169 

This product is called the magnetic moment, M, of the magnet. 
(Give a physical definition of the term magnetic moment). The 
expression for the torque acting on a magnet when placed in a 
uniform field therefore is 


L = MH sin 6 170 

where M = ml. Using the method employed in the case of the 







THE ELECTRIC CURRENT 


127 


gravity pendulum we may show that a magnet set vibrating in a 
magnetic field with small amplitude executes simple harmonic 
motion and that the expression for the period, is 

T = STrVf/MH 171 

I is the moment of inertia of the magnet about the axis of sus- 
pension. 

158. Magnetic Potential. By analogy with the electrostatic 
case the student should define magnetic potential, unit magnetic 
potential and potential difference and show that the potential 
at a distance, r, from a pole, m, is 

y A3. 172 

mi- 

problem 

Calculate the value of the field due to a magnet for the following 
positions (a) on the axis of the magnet a distance, s, from the 
center of the magnet; (b) on the perpendicular bisector of the 
line joining the poles a distance, s, from the center. 

THE ELECTRIC CURRENT AND ITS EFFECTS 

159. Production and Detection of Electric Current. Electric 
charges in motion constitute an electric current. An electric 
current may be produced by an electrostatic machine, voltaic 
cell, thermojunction, by the movement of a conductor across a 
magnetic field or by various other means. No matter by what 
means the current is produced its effects are the same. Electric 
current may be detected and measured by the heating, chemical 
or magnetic effect. The magnetic effect lends itself especially 
well to quantitative measurement and for that reason will occupy 
our particular attention. This effect may be observed by simply 
placing a compass above or below a wire carrying a current, 
provided the wire is held in a north and south direction. The 
deflection is evidence of a magnetic field associated with the 
current flowing in the wire. Experiment shows that this magnetic 
field may be represented by lines of force drawn as concentric 
circles with the wire at the center. If the wire is grasped with 
the right hand in such a way that the thumb points in the direction 
in which the current flows the- fingers will indicate the direction 
of the field. 

160. Side Push on a Wire Carrying Current. A wire placed 
at right angles to a magnetic field is acted upon by a force tending 



128 


ELECTRICITY 


to move the wire sideways in 
a direction at right angles to 
the magnetic field and toward 
the weaker part of the field. 

In Fig. 101 P indicates the 
cross-section of a vertical wire 
in a horizontal field, H 0 . The 
field due to the current in the 
wire is shown a’s a circle around 
the wire, and will be designated 
in this text by the symbol H 
Fig- 101 without the subscript. In this 

case the current is flowing down into the plane of the paper. The 
field is weakened below the wire and the force on the wire is 
downward. A wire parallel to a magnetic field experiences no 

force. If the wire makes an 




angle, 9, with the field (Fig. 

. 102) the force is merely that on 
the component,/sin, 9 perpen¬ 
dicular to the field and is 
normal to the field. 

Experiment shows that the 
force acting on a straight 
wire making an angle, 9, with 
a magnetic field depends upon 
the current, the field, the 
medium, the length of wire and the angle, in the manner given by 
the expression 

f = k i n H 0 Z sin 6 173 

where k is a proportionality constant dependent upon the 
units only. This equation may also be written as 

f = k i B Z sin 6 173 

where B is the magnetic induction. 


161. Magnetic Field due to Electric Current. Experiment 
shows that the magnetic field due to a current in a wire may be 
expressed by the relation 


dH 


k i dZ sin 6 
~a 2 


174 


where i is the current; dZ, an element of length along the wire; 
9, the angle between dl and the line drawn from dl to the point 
at which the field is being measured; a, the distance from dZ to 
this point; k, a proportionality constant determined by the units 
in which i is measured. 



















THE ELECTRIC CURRENT 


129 


We may derive this equation from 173, assuming that equation 
as an expression of the experimental facts of the case. Imagine 


a test pole, m, placed at o, 
Fig. 103. The field at dl 

due to this test pole is-^V. 

;ua 2 



If this is substituted in eq. 
173 for H 0 we have 

df = k i d l T ^r sin 0. 
a J 

The reaction on the test pole 
is mdH, where dH is the 
field at o due to current 
flowing in dl. Therefore; 
equating action and reaction 
and cancelling m we have 
JXJ _ k i dl sin 6 

Ap- 

turns we note that a is the 


162. Magnetic Field at the Center of a Circular Loop. 

plying eq. 174 to a loop of N 


radius and 0 is—. Hence the equation becomes 


dH 


kNidl 

r 2 


or 



k 27r Ni 
r 2 


175 


163. The Electromagnetic c. g. s. Unit of Current and the 
Ampere. By means of equation 173 we may define unit current 
as that current which will cause a wire to experience a side push 
of one dyne per cm. length when placed at right angles to unit 
magnetic field in air. Unit current is more often defined in 
terms of the field due to a current in a circular loop. It has 
been agreed to say that unit current flows when the field at the 
center of a circular loop of one turn and unit radius is 2t dynes. 
This is equivalent to the first definition. With the unit of current 
so defined k of eq. 175 becomes unity. 


The ampere, the practical unit of current, is 1/10 of the c. g. s. 
unit. (When current is expressed in amperes the symbol, I, will 
be used. All c. g. s. units will be indicated with small letters 
practical units with capitals). 


164. The Field on the Axis of a Circular Coil, a distance, p, 
from the center is 







130 


ELECTRICITY 


H = 


27rNir 2 


176 


(r 2 + p 2 ) 3 / 2 . 

The field at o due to the element dl is 

dH = . Observe that 6 = and that the component of 

dH perpendicular to p will be neutralized by a component from 
the element ddiametrically opposite dl, Fig. 104. To get 
the resultant field at o, it is necessary then to consider merely 



the component along p due to dl and integrate the expression 
for the whole coil. 


H = 



id l sin a 
a 2 


This leads to equation 176. 

165. The Field near a Long Straight Wire carrying current is 
H '= 2i/r 177 



By LaPlace’s equation 
the field at o due to dl is 

jrT i dl sin d 

ail =--- 

a 2 

If all variables are ex¬ 
pressed in terms of the 
angle </> (thus sin 6 = 
cos <f), ad <f> = dl cos </> and 
a = r/cos 0) and the re¬ 
sulting expression inte¬ 
grated between the limits 
</> = 7r/2 and 0 = — tt/2. 
We get equation 177. 


Fig. 105 













166. 


THE ELECTRIC CURRENT 

Field at Center of a Long Solenoid is 

47rNi 


H = 


131 


178 


(N being the total number of turns.) 

The figure shows a longitudinal section of a solenoid. The 
field at o due to length of solenoid dL is by equation 176 
_.jj _ 27rNd/ir 2 
d “ l (r 2 + P 2 ) 3 / 2 

where N is the number of turns. This may be written dH = 

27rNir 2 d//a 3 . Note 
that ad0 /dl = sin </> 
and that r/a = sin 0* 
Making these substitu¬ 
tions, and integrating 
between the proper limits 
equation 178 is ob¬ 
tained. 



<§)^>#(§)(§)<§)(§)(§)(§)<§>(§)<!)(§)##(§)# 167. Work to Carry 

Unit Pole around a Wire 


Fig. 106 

Carrying Current. Using equation 177 one may show that the 
work done in carrying a unit magnetic pole once around a wire 
carrying current, i, is 

W = 47ri 179 


It may be noted that this work is independent of the path followed. 

168. Field at the Center of a long Solenoid. Alternative 



Derivation. In the case of 
a circular solenoid, if we 
may assume that the work 
required to carry a unit 
pole once around any path 
enclosing any number of 
turns of wire, each carrying 
current, i, is 47r i times the 
number of turns (eq. 179), 
and if we traverse any path 
around the solenoid, such 
as the dotted line, Fig. 
107, we may write the ex¬ 
pression 


Fig. 107 














132 


ELECTRICITY 


H * 27nr = 47ri * 27rrN 

where N is the number of turns per cm. length. 
This gives H = 47rNi 


or 


H = 


47rNI a 
10 


If N is taken to mean the total number of turns the expression 
takes the form 


47rNI 

H ior" 

A long straight solenoid is simply a special case of the above, 
the radius then being infinite. 

180. Quantity of Electricity. Unit Quantity. The product of 
current and time gives quantity of electricity. The amount of 
electricity flowing by a point in one second when the current is 
one ampere is the practical unit of quantity and is called the 
eoulomb. Quantity is expressed by the equation 


Q - / I d t 


179 


MEASUREMENT OF CURRENT 


181. The Tangent Galvanometer. If we place a compass 

needle at the center of a 
coil carrying current and the 
coil is arranged so that its 
field is perpendicular to the 
earth’s field, the deflection of 
the compass needle may be 
used to measure current. 
Such a device is called a 
tangent galvanometer. The 
relative directions of the field 
due to the earth and that due to 
the current, and their resultant are shown in the diagram. The 
compass needle will place itself parallel to the resultant, de¬ 
flecting through an angle, 0, from its original north-pointing 

position. With equation 175 in the form, H = 2 ,and the 

relation,-^— = tan 0, we may establish the relation, 
tl e 

I = 2 7rN e tan ^ = tan 0 
I 0 is called the constant of the galvanometer. 



180 








MEASUREMENT OF CURRENT 


133 


The sensibility (or sensitiveness) of a galvanometer is often 
expressed in terms of the current per unit scale division deflection. 
The smaller this current is, the greater the sensibility. Equation 
180 shows upon what factors the sensibility of a tangent galva¬ 
nometer depends. 

In the case of the D’Arsonval galvanometer (see paragraph 
183) sensibility is also often expressed in meg-ohms, that is in the 
number of million ohms which must be placed in series with the 
galvanometer to give a deflection of 1 mm. on a scale 1 meter 
distant, when the e.m.f. in the circuit is 1 volt. 

182. Torque on a Coil Suspended in a Magnetic Field. In 
the case of a rectangular coil suspended in a magnetic field, H, in 
such a way that the plane of the coil makes an angle, 8, with the 
direction of the field, it may be shown that the torque, L, is 


L = 


NAIH 

10 


cos 8 


181 




A is the area of the coil; n is 
the number of turns; H is the 
field in which the coil is sus¬ 
pended. The figure shows 
both plan and elevation of 
the coil as suspended in the 
magnetic field. The side 
push on each vertical oortion 
NIjjlHZ 


of the coil is 


10 


since /x is unity for air, 
The value of 


or, 


NIHZ. 


10 

the turning 

i * • NIHZ 

couple or torque is 


ab cos5 or L = —cos 8 

183. The D’Arsonval Galv- 
nometer. A coil suspended 
in a magnetic field as de¬ 
scribed above constitutes a 
device for measuring current. 
A D’Arsonval galvanometer 
consists merely of such a 
movable coil suspended in a 
strong permanent field. 

The suspension is so ar¬ 
ranged that the plane of the 




























ELECTRICITY 


134 


coil is held parallel to the field when no current is flowing. When 
current flows, the magnetic torque expressed by 181 tends to 
turn the coil to a position normal to the field. Thereby the 
suspension is twisted and an elastic' return torque, introduced. 
Measured in terms of the angle of twist, 8, the elastic torque is 
L 0 5. 

When the coil is at rest this is equal to the magnetic torque (eq. 
181). Equating these and solving for I we have 

10 L 0 _8_ 

1 NAH cos 8 


and for small angles 


I = 


10 L 0 
NAH 


tan 8 


If we measure tan 8 by means of deflection, s, on a straight 
scale at some fixed distance, d, from the coil we have for small 


s 

angles tan 8 = which we may substitute in the above. 

If the indication is by means of a ray of light reflected from a 


mirror fastened to the coil then tan 8 


— and the equation be¬ 


comes 


= JOL^ 
2NAHd 


182 


. In the case of a radial field the student may show that the 
equation becomes 


I = 


10 L 0 
NAH 


k 8 


183 


where 8 is the angle of deflection of the coil. The equation 
shows upon what factors the sensibility of a D’Arsonval galva¬ 
nometer depends. 


OHM’S LAW 


184. Joule’s Law; Resistance. Experiment shows that the 
heat developed in a conductor by an electric current is proportional 
to the time, to the square of the current, and to a factor, R, 
called the resistance which depends upon the material and di¬ 
mensions of the conductor. This is expressed by the following 
equation known as Joule’s law, namely, 

Hj = kRI 2 t 184 

Hj is the heat expressed in mechanical units; k is a proportion¬ 
ality constant and depends merely upon the units chosen. Like¬ 
wise experiments show that R depends upon the dimensions in 
the following way, namely, 







OHM’S LAW 


135 


R = — 185 

a 

where / is the length; a, the cross section; p, a constant depend¬ 
ing upon the material and temperature, p is called the specific 
resistance. The reciprocal of p is called the electrical conductiv¬ 
ity. 

If the conductor is of such dimensions and material that a. 
current of one ampere develops one joule of heat in one second,, 
the resistance, R, is then said to be one ohm. This defines the 
ohm as the practical unit of resistance. With this choice of 
units k (eq. 184) becomes unity and need not be expressed. 

The c. g. s. unit of resistance is similarly defined as such a 
resistance that one c. g. s. unit of current develops one erg of heat 
in one second. It follows that one ohm equals 10 9 c. g. s. units. 

It is permissible to express H in equation 184 in calories, in 
which case the equation becomes JH = RI 2 t, J being the me¬ 
chanical equivalent of heat. 

185. Ohm’s Law for a Portion of a Circuit. The development 
of heat in a circuit when current flows means the consumption of 
energy or the doing of work against the resistance to the flow 
of electricity. The difference of potential , Pd, between two points 
in a circuit is defined as the work required to carry unit quantity 
of electricity from one point to the other. From this definition 
we may write Hj = PdQ, or since Q = It 

Hj = Pdlt 186 

Combining equations 184 and 185 we have 

Pd = RI 187 

This is Ohm's Law as applied to a portion of a circuit in which 
energy is used as heat only. 

We have a potential difference of one volt between two points 
when the current is one ampere and the resistance is one ohm. 
The c.g.s. unit of potential difference is defined similarly. 
The units of pd. may also be defined in terms of work by means 
of equation 186. Thus the Pd is one volt if the heat developed is 
one joule per coulomb; or it is one c.g.s. unit, if the heat is one erg 
per c.g.s. unit of quantity. 

186. Ohm’s Law for a Complete Circuit. In the case of a com¬ 
plete circuit the energy consumed or work done in carrying unit 
quantity of electricity completely around the circuit is necessarily 
the energy supplied to the circuit per unit of quantity and is called 
the electromotive force of the circuit. This energy usually 
appears either as the result of chemical reactions in a voltaic cell, 
or at the expense of mechanical work in a dynamo. For a complete 
circuit then we may write 

E = (R + r)I 188 

E is the energy supplied in the circuit per unit quantity of 


136 


ELECTRICITY 


electricity or in other words the electromotive force of the cells 
or generators in the circuit. In the case of a cell it is numerically 
the difference of potential across the terminals on open circuit, 
r is the resistance within the generator. In case the circuit 
contains several generators, E is the algebraic sum of all the 
e.m.f.’s. 

187. General Expression for Ohm’s Law. In general any e. 
m. f. is considered positive if it aids the flow of current, negative if 
it opposes, if we consider the direction in which the current flows 
around the circuit as the positive direction. In the case of a net¬ 
work or mesh the current may flow in different directions in 
different parts of the mesh. It is then necessary to assume a posi¬ 
tive direction and consider that E, which stands for any cell or 
generator which may happen to be in any particular branch, is 
positive if it tends to send current in the assumed positive direction 
and RI positive if I flows in the positive direction. If E 
stands for the energy supplied between two points in the circuit 
per unit quantity of electricity passing, and RI for energy 
consumed or converted to heat per unit quantity, then the differ¬ 
ence between these two quantities gives the Pd or the net work 
performed in carrying unit quantity of electricity between the 
points in question. Therefore we may write 

Pd = RI — E or Pd = SRI — 2E 189 
This equation is the most general expression of Ohm’s law. 

The special cases of Ohm’s law may all be obtained from this 
general expression. Thus, if there is no e. m. f. between the points 
in question E = 0 and we have Pd = RI. If we have an e.m.f. 
between the points but no current flowing (open circuit) then 
RI = 0 and Pd = — E. If we go completely around a circuit or 
closed mesh Pd = 0 and we have 0 = RI — E or E = RI or 
2E = SRI. 

188. Kirchhoff’s Laws. We have just shown that in going 
completely around any circuit or closed mesh 

S = SRI 190 

This may be applied to any closed circuit or network, or mesh, 
of even a very complicated circuit. The conventions regarding 
signs must be carefully observed. Another relation is obtained 
by equating the sum of all the currents flowing up to a point to 
zero. That is 

S I = 0 191 

This simply states the fact that there is no accumulation of 
electricity, that is, as much flows away from a point as flows 
up to it. (This statement is not true for variable currents in 


OHM’S LAW 


137 


circuits containing capacity. The law in that case must be 
modified.) Equations 190 and 191 are known as Kirchhoff’s 
laws. In applying these laws the procedure may be summarized 
as follows: 

1) Assume a direction, say clockwise, around the mesh as posi¬ 
tive. 

2) Assume a direction for the current in each branch. If 
direction is wrong, current will come out negative. 

3) Equate the algebraic sum of the e. m. f s. to the algebraic 
sum of the RI terms, paying proper regard to signs. 

4) Equate the algebraic sum of the I’s at one of the junctions 
to zero. 


An equation according to 190 and one according to 191 may 
thus be written for each mesh. Each equation must contain one 
term not found in any of the others. There may thus be written 
as many equations as there are unknown quantities. 


189. Application of Ohm’s Law to a Simple Circuit. To 


I—WMA 
v 2 


d 


A 

b^AMMA 



pM/W\M/W\A 

B 



Fig. 110 


illustrate the method 
of applying Ohm’s 
law for the various 
cases that arise, let 
us consider the circuit 
of Fig. 110. Cell A 
has an e. m. f. of 2 
volts and a resistance 
of 2 ohms; cell B an 
e. m. f. of 5 volts and 
a resistance of one 
ohm. The direction 


of the current is clockwise, since B has the larger e. m. f. The 
value of the current is found from the equation, 2E = SRI, which 
gives I = H ampere. If we start at (a) and travel clockwise around 
the circuit step by step, applying Ohm’s law in its general form, 
Pd = RI — E, we have 


apdb =1X^-0 = + ^ 


b Pd c = 1 X y 3 - (+5) = -4^ 

cPd d = 2 X H— 0 - + % . 

d Pd e = 2 X y — (—2) = +2H 

ePda = 3 X H — 0=+l 

SPd= + M — 4%+^+2M+ 1=0 












138 


ELECTRICITY 


MEASUREMENT OF RESISTANCE AND E. M. F. 


190. The Wheatstone Bridge. 


If an unknown resistance is con¬ 
nected as shown 
with three other 
resistances, a 
galvanometer 
and cell, the cora¬ 
ls bination is called 
a Wheatstone 
Bridge. If the 
resistances rj, r 2 
and R are so 
adjusted that 
no current flows 
through the gal¬ 
vanometer the 


c 



value of x may be computed. Under these circumstances 
points c and d are at the same potential. By applying Ohm’s 
law we may establish the necessary relation viz., n/r 2 = x/R. 

The student may show that galvanometer and battery may be 
interchanged without affecting the balance. 

191. The Potentiometer. In the circuit shown a b represents 

a uniform resist¬ 
ance such as a 
slide wire, S is a 
standard cell of 
known e. m. f., X 
a cell whose e. m. f. 
is desired. If the 
key, k, is closed 
on m and the 
position of c ad¬ 
justed so that no 
current flows 
through the gah 
vanometer (it is 
of course necessary 
that S be so di¬ 
rected as to op¬ 
pose the cells E) the pd from a to c must equal the e. m. f. of cell 
S. Likewise if the key, k, is closed on n and the point of 
balance, c', found the pd from a to c' must equal the e. m. f. of the 
unknown cell. We have the relations 
ac-pl = E and ac'pl* = E x 
















MEASUREMENT OF RESISTANCE AND E. M. F. 


139 


where p is the specific resistance. By combining these 
equations we have 

ac _ E s 
ac' Ex * 


This device is called a potentiometer. We may use it to 
measure current by measuring the pd across a known resistance, 
or to measure resistance by comparing the pd across a known 
resistance with that across the one whose value is desired, the 
two resistances being in series in an auxiliary circuit. 

192 Voltmeters and Ammeters. A voltmeter is a high re¬ 
sistance galvanometer used to determine potential difference be¬ 
tween points in a circuit. It is always shunted around the portion 
of the circuit between the points in question. Its resistance 
should be such that the current diverted through it is negligible as 
compared with the total current. An ammeter is a low resistance 
galvanometer used to measure current. It is always connected 
directly in the circuit. Its resistance must be so small that the 
insertion of the instrument in the circuit produces a negligible 
effect on the resistance of the circuit. 

193. Resistances in Parallel. The figure shows a portion of a 
circuit in which current, I, is flowing across a number of resistances 

ri, r 2 , r 3 connect- 



Fig. 113 


ed in parallel. By 
observing that the 
sum of all the 
currents Ii + h 
+ I 3 + • • • = I 
and applying 

Ohm’s law, it may 
be shown that 
that the equiva¬ 
lent resistance, 
R e , is given by 
the expression, 

1 /R e = 1 /n + 1 /r 2 + l /r 3 + • • • 192 

194. Effect of Temperature on Resistance. In the case of 
metals the resistance is found experimentally to increase with 
temperature according to the following law, 

R t = Ro(l -f~ at + /3t 2 -)-•••) 193 

/3t 2 and the terms following are usually negligible, a is called 
the temperature coefficient of resistance. 

In the case of the weak electro-positive elements such as carbon, 
silicon, tellurium, etc. and in the case of most oxides and sulphides 
a negative temperature coefficient is found, the resistance decreas¬ 
ing according to an exponential law. 

In the case of electrolytes, both liquid and solid, a negative 





140 


ELECTRICITY 


temperature coefficient is also found. The resistance change 
here also follows an exponential law but of a somewhat different 
type than in the preceding case. 

195. Power and Energy in the Electric Circuit. Power is the 
time rate of doing work. Therefore in the electrical case PdQ /t = 
Power. This may be written as 

Power = Pd I 194 

Power is here expressed in joules per second or watts. 

Energy is given by the relation Pd'Q or PdTt. Energy is 
commonly expressed in kilowatt-hours, the unit being the energy 
consumed per hour when used at the rate of 1,000 watts. 


INDUCED ELECTROMOTIVE FORCES 

196. Lenz’s Law. We owe to Faraday the first experimental 
demonstration that electric current will flow in a closed circuit 
whenever any portion of such a circuit is moved across a magnetic 
field or whenever a magnetic field in the neighborhood of such a 
circuit is caused to change. Such currents are called “induced 
currents” and the corresponding electromotive forces are called 
“induced electromotive forces.” Induced electromotive forces 
are found by experiment to be proportional to the rate of cutting 
lines of force or the rate of change of magnetic flux through the 
circuit. We have seen that a conductor carrying current, i, and 
placed at right angles to a magnetic field, H, experiences a force, 
i / H. Such a force will be experienced whether the conductor 
is at rest or is moving relative to the field. In case the current 
flowing is that induced by the motion of the conductor across the 
field, the force experienced by the conductor is found to be in 
such a direction as to oppose the motion, and is given by the 
expression, —i / H. In moving against this force with velocity, 
v, work must be done at the rate, —i l Hv. Electrical energy 
appears in the circuit at the rate, e i, (expressed in c.g.s. units). 
The electrical energy appearing per second must equal the me¬ 
chanical work done per second. That is, 

. 7 tt d0 195 

ei = —i l Hv or e --^ 

at 

Where = l H v, the rate of cutting lines of force or the rate of 

change of flux through the circuit. 

If practical units are used, E I expresses power in watts and the 

. , T7T I l Hv 

expression becomes EI =-• 

/Hv _— d 0 
W* “ 10 8 dt 

one conductor the induced e. m. f. is 


Or E = 


If there is more than 




INDUCED ELECTROMOTIVE FORCES 


141 

196 


Nd <f> 
e dt 

where N is the number of turns or conductors cutting the field. 
This is known as Lenz’s Law. Expressed in volts 

w _ Nd 0 197 

^ 10 8 dt 

By use of Ohm’s law and equation 197 we may find the ex¬ 
pressions for current and for quantity of electricity induced. 

197. Alternating Current Dynamo. If a coil is set to rotating 
in a uniform magnetic field the 
e. m. f. induced at any instant may be 
shown to be 

27rNn<£ m sin 0 198 

D = -—rr- 



10 8 

It should be noted that the flux 
through the coil at any instant is 
</> cos 0. 

The diagram shows a plan of such 
a rotating coil. <t> is the flux 

through the coil when at right angles to the field (maximum flux); 
6 is the angle between the field and the normal to the plane of 
the coil or the angle the coil has turned through from its position 
of maximum flux; N is the number of turns in the coil, n the 
number of revolutions per second. Equation 198 may be de¬ 
rived by applying equation 197, thus 

p N d <f> _ N (f>m d cos 6 _ N0 m sin0d0 

h 10 8 dt W dt “ 10 8 dt 

Now 0 = cot = 27r n t, where co ( = 27m) is the angular velocity. 


d0 0 

— = co = 27rn. 
at 


By making these substitutions 198 is obtained. 


Equation 198 may be written in the form 

E = E 0 sin cot 199 

where E 0 is the maximum value of E and is given by the ex- 
27rNn <f> m , 


pression E 0 = 


10 8 


It should be noted that the e. m. f. is maximum when the plane 
of the coil is parallel to the field and zero when at right angles. 














142 


ELECTRICITY 



If the ends of 
the coil are ar¬ 
ranged to make 
contact on slip 
rings as shown, 
Fig. 115, and con¬ 
nection is made 
to an external 
circuit from the 
rings, a current 
will flow in this 
circuit whose 
value alternates 


from a positive to a negative maximum and back to positive 
once every cycle. In other words we have an alternating current 
generator. If we have a second turn or winding at right angles 
to the first and connected to a separate pair of slip, rings and an 
independent external circuit we have a two phase generator, 
the current in one circuit being a maximum when that in the other 
is zero. Such currents are said to be 90° apart in phase. In 
practice it is usual to use only three slip rings, the two windings 
having one ring in common. The external circuits have a common 
return wire connecting to the common ring and taken together 
constitute a three wire system. We also have three-phase and 
polyphase systems the discussion of which at this point would 
lead us too far astray. In the discussion so far we have assumed 
a uniform field such as would exist between the pole faces of a 
ring magnet, the cases under discussion being those of a two pole 
generator. Generators with four, six, eight, and sometimes 
more poles are used in which cases the current reverses between 
every pair of poles. Thus with an eight pole generator the 
current reverses four times a cycle. 


198. Direct Current Dynamo. If at the instant the e. m. f. 
reverses (or is at its zero value) the connections to the external 
circuit are reversed, the current in the external circuit will always 
be in the same direction. Its value will vary, somewhat as indi¬ 
cated in the curve, Fig. 116 (a). If a winding at right angles is 
placed in series with the first, the circuit will possess two e. m. f’s., 
the resultant being indicated by the dotted line Fig. 116 (b). 


Fig. 116(a) 













INDUCED ELECTROMOTIVE FORCES 


143 



Fig. 116(b) 

We thus have a fluctuating direct current, the fluctuations of 
which may be smoothed out by connecting in series a sufficient 
number of properly spaced armature windings. 

In case the field is radial the rate of cutting lines of force by a 
given turn or set of turns is constant for each half revolution. 
The coil passes abruptly from the space before the (+) pole to 
that before the (—) pole and the direction of the e. m. f. abruptly 
changes. If at this instant the external connections are reversed 
we have then an interrupted direct current in the external circuit 
of the form shown in Fig. 116 (c). The e. m. f’s., of this sort 

Tnry~r~r 

Fig. 116(c) 

from the successive turns or sets of turns, when connected in 
series, give a steady direct current. 

There are two types of armatures used in direct current ma¬ 
chines, one known as the Gramme ring, the other as the drum 

armature. Theformer 
consists of an iron ring 
with a continuous wind¬ 
ing, taps from the different 
turns being led out to com¬ 
mutator bars, Fig. 117 (a). 
The bars pass under 
brushes at the top and 
bottom (the neutral posi¬ 
tions). It will be noticed 
that only the outer part 
of the turn cuts lines of 
force, there being no field 
within the ring, and that 
for clock-wise rotation the 
current flows toward the 
bottom brush from the 
windings on each side of 
the neutral plane (vertical 
plane through the axis). 



















144 


ELECTR CITY 


The turns on either side of the neutral plane are thus seen to 
be in parallel. The e. m. f. may be obtained by applying equation 
196. Thus, taking N as the number of cutting conductors, in 
this case equal to the number of turns, we have 

E = (^-) But v = co r = 27rnr, hence 

2 10 8 


N7H.27rnr 
" 2xl0 8 ’ 


Since the field is radial the total flux, </>, is given 


by the expression </> = tt l H. 

Therefore E = 200 

10 8 


In the case of a drum armature, the windings are placed longi¬ 
tudinally on an iron cylinder. The windings are in series and taps 
are brought from successive turns to the commutator bars. The 
diagram shows a four commutator drum armature. The field is 
radial. With a clockwise rotation the current would flow as 
indicated by the arrows through two paths from the lower brush 



to the upper one. At any instant one-half the wires are in paral¬ 
lel with the other half. As before E = ^ N is the 

number of cutting conductors but in this case not the number of 
turns but double that number. By the same reasoning as in the 

preceding case the e.m.f. is given by the expression E = t L^L£ 

10 8 

This expression requires that there be at least one pair of turns, 
i. e., four cutting conductors. In practice there are many turns. 

198. The Direct Current Motor. When a current, I, flows in 
a conductor which is at right angles to a magnetic field the con- 





















INDUCED ELECTROMOTIVE FORCES 145 


ductor will be impelled to move under the action of the force 

which we know to be expressed by the equation, f = —. 

As a result of the movement of the conductor across the field 
a current, I', will be induced in the opposite direction and an 
opposing force f' (= —I' l H) will be developed. The conductor 
will gain velocity until I' = I, (frictionbeing considered negligible). 
Thereafter the velocity will remain constant. In this case no 
work is being done since the resultant current and therefore the 
resultant force is zero. In case there are retarding forces against 
which energv has to be expended then the current does not drop to 

zero. The force exerted then is f = where I is the re¬ 


sultant current, and work is done at the rate ergs P er 

N I l Hv . N Z H v . , . f 

sec., or—— joules per sec. —— is the expression for an 


10 8 


e. m. f., the opposing e. m. f., of the motor, E m . Therefore 
electrical energv is being converted to work at the rate, E m I. 

The direct current generator with Gramme ring or drum arma¬ 
ture would constitute a motor if direct current were sent into the 
armature windings and would exhibit the phenomena described 
above, the motion being of course that of rotation. The energy 
equation for a direct current motor is 

E g I = R I 2 + E m I 201 

Where E g is the e. m. f., of the generator; E m , the back or 
counter e. m. f., of the motor; and R, the resistance of the circuit. 

E m I = ^ where f is the force described above. This term 

represents the rate at which mechanical work is done. R I 2 
is the rate at which energy is dissipated electrically as heat. 
199. Efficiency of a Motor (with reference to the conversion 

of electrical energv to work). Efficiencv = the 

case of the motor, 

Efficiencv = = vr- 202 

.EgI g 


PROBLEM 

Show that a motor works at maximum rate when the counter 
e. m. f. is one half the impressed e. m. f., that is, when the efficiencv 
is 50 per cent. 







146 


ELECTRICITY 


Note that the mechanical work depends upon the back e. m. f. of the motor 
and is given by the expression EmI; and that when the mechanical work is a 
maximum, its rate of change with Em must be zero. Differentiate equation 
201 with respect to Em and place 

dEm (EmI) = °- 

The truth of the statement of the problem will then appear on inspection of 
the result. 


200. The Induction Motor. If a magnetic field is caused to 
move across a conductor which forms part of a closed circuit, 
the conductor will, as we have seen, experience a force opposing the 



relative motion between it and 
the field. If free to move the 
conductor will be dragged a- 
long with the field. It is 
possible by the arrangement 
shown in Fig. 118 to pro¬ 
duce a rotating magnetic 
field. A coil placed sym¬ 
metrically in such a field 
would be impelled to rotate 
with the field. 


The rotating field is produced 
by sending alternating current 
I 0 sin 27rnt through A and B and 
current I 0 cos 27rnt through 


C and D. If we have a two-phase circuit and connect one phase 
to A and B and the other to C and D we will have the conditions 


desired, since the currents are then given by I 0 sin 27mt and 
I 0 cos 27rnt. Now if we have a uniformly rotating vector of value, 
H 0 , its projection on the vertical diameter at time, t, is Hi = 
H 0 sin 27rnt, and on the horizontal diameter H 2 = H 0 cos 27rnt. 
Therefore we may represent our two fields as the projections on 
vertical and horizontal diameters of a rotating vector, H 0 . 


The resultant field has the value at any instant of 


VHi 2 + H 2 2 or H 0 V sin 2 27rnt+ cos 2 27rnt, which is H 0 since 


V sin 2 27rnt+ cos 2 27rnt = 1. Thus the composition of the two 
fields gives a rotating field of constant value, H 0 , which is 
the maximum value of either field. 









147 


SELF AND MUTUAL INDUCTION 


The movement of this field across the coil produces current. 



Fig. 119 


F, in the coil. If the coil is 
rectangular the corresponding force 


per turn is 


21 l H 
10 ' 


If there are 


N turns and if the radius of the 
coil is r, we have the torque, 
2 N V l H r 

-Jq -, under which the coil 


rotates. The direction of this 
torque will be such as to oppose the 
relative motion between the coil 
and the field, in other words, to 
rotate the coil with the field. 


SELF AND MUTUAL INDUCTION 


201. Self Induction. If the current through any circuit 
changes the magnetic flux through the circuit will change and an 
induced e.m.f. will thereby be developed which by Lenz’s Law 
will oppose the change in flux. The e. m. f. will be given by the 

expression E = — an< ^ s ^ nce the °f change of flux is 


proportional to the rate of change of current we may write this in 
the form 


E = — L *= 203 

dt 

where L is the proportionality constant. L is called the 
coefficient of self-induction or the inductance of the circuit. It 
is numerically equal to the e. m. f. in volts induced when the 
current changes at the rate of one ampere per second. If the di¬ 
mensions of the circuit are such that an e.m.f. of one volt is thus 
induced the inductance is said to be one “henry.” 

In the very important case of a ring solenoid or a straight 
solenoid which is long compared with its diameter we may readily 
compute the inductance. Eq. 197 may be written as 

= L an d a l so as d 4> = Ld I. If this is integrated be- 
llrdt dt 1U B 


tween the limits 0, and d>, and 0, and I, we have ^ = LI. 

— 1U 8 

</> = aB = a ^ H. Where B is the magnetic induction; /jl, the perme¬ 
ability of the medium; and H, the field. For either case under 








148 

consideration 


ELECTRICITY 


H = 


47m I 


10 / 

Therefore L = 


Thus 0 = 


47rn la n 
10 / 


47rn 2 a /jl 
M" 


204 


be noted. If the 
jji may be equal to 


The importance of the factor, /x, should 
solenoid happens to be wound on an iron core, 

1000 or more. 

202. Mutual Induction If the flux from one coil or circuit 
threads wholly or partly through another circuit any variation 
in that flux will produce an induced e.m.f. in the second circuit. 
This e.m.f. will be proportional to the rate of change of flux and 
therefore to the rate of change of the primary current, and may be 
expressed bv the equation 
dip 


E s = 


M- 


dt 


205 


dL 


E s is the e.m.f. in the secondary circuit, -- j s the rate of 

change of current in the primary circuit. M, the proportionality 
constant, is called the coefficient of mutual induction and is 
defined in the same way as L and measured in the same units. 

In the important case of a primary and secondary coil wound 
on an iron ring, in which case the flux from the primary all threads 
through the secondary we may get an expression for M as follows: 
Equation 205 may be written 

N s d 0s , , dL 


= M 


where N s is the number of 


10 8 dt dt 

turns in the secondary and 0 S is the flux through the secondary. 
In this case 0 S is also equal to 0 P , the flux through the primary. 
We may therefore write 


N s 0p 
10 8 

0 P — a fj, H — 


= M I p 
47 tN p I p a/x 


hence 


10 / 

= M 206 

203. Ohm’s Law for a Circuit Containing Inductance. In 

this case we have merely the additional and opposing e.m.f., 
dl 

L —. Hence the equation is 


dt * 


E 


L d J = 

L dt 


R I 


207 


When current is made to flow against an induced e.m.f., 
Ei, the additional energy is given by the expression, 











THE BALLISTIC GALVANOMETER 


149 


W = J E x Idt 

If the applied e.m.f. is E, the energy input is J*EIdt while that 

appearing electrically is only J (E—Ei) Idt. The difference, 

namely J^EiIdt, is the energy stored in the magnetic field. If 

Ei is the e.m.f. of self induction the student may show that the 
energy of the magnetic field is 

W = l LI 2 208 

Comparison of this expression with the formula for the kinetic 
energy of a moving body namely W = 1/2 mv 2 brings out an 
interesting and useful analogy between inductance and inertia. 
In the effects of inductance the analogy is remarkably borne out. 

THE BALLISTIC GALVANOMETER AND MEASURE¬ 
MENT OF CAPACITY 

204. The Ballistic Galvanometer. A galvanometer (either 
D’Arson val or tangent) may be used to measure quantity of electric¬ 
ity by means of the throw or kick:of the coil or needle, if the time 
during which the electricity is passing is small compared with 
the period of the galvanometer. For example, if a galvanometer 
is in series with a condenser the momentary charging and dis¬ 
charging current must pass through the instrument. The time 
of passage is extremely small, the coil or needle receiving simply 
an impulse which causes it to swing out from its rest position. 
It immediately swings back and usually after one or more vibra¬ 
tions comes to rest. 

A change in the magnetic flux threading a circuit will produce 
a momentary current. If the change is a sudden one such that 
the whole quantity of electricity induced passes through the 
galvanometer in a short time compared with the galvanometer 
period, we get here also a simple throw which is proportional to 
the quantity of electricity which passes (see below). 

Whether the momentary current is the direct charging current 
of a condenser or an induced current due to a sudden change of 
magnetic flux', the effect on the galvanometer will be a throw of 
the coil or needle. The following discussion (which refers to a 
D’Arsonal galvanometer) shows that the quantity of electricity 
may be measured by the throw and indicates exactly what factors 
are involved. A galvanometer used in this way is called a “ballis¬ 
tic galvanometer.” 

205. The Equation for the D’Arsonal Galvanometer when 
used Ballistically. Assume that the total quantity of electricity 


150 


ELECTRICITY 


passes through the galvanometer before the coil moves appreciably 
from rest and for simplicity assume an approximately radial field. 
The magnetic torque at any instant is 


j _ IHaN 
L 10 


(see eq. 181) 


The impulse of the torque is 


J Ldt ■ J 


Idt HaN 
10 


NaHQ 

10 


since Jldt = Q. The impulse of the torque is equal to the angu¬ 
lar momentum, hence 

Kco„ = 209 


K is the moment of inertia of the moving part; co 0 , the initial 
(maximum) angular velocity; a, the area of the coil; N, the num¬ 
ber of turns; H, the field; and Q, the quantity of electricity 
to be measured 


The kinetic energy at the start is- Kwq. At the end of the 

swing this has been converted to potential energy of the twisted 
suspension, viz, 


where L 0 is the constant of torsion of the suspending wire and 
d Q is the maximum throw in radians. Hence 

2 Kcoo = ^ Lo 5q or Ko?o = Lo 5o 210 

Combining equations 209 and 210 we get 

NaHQ = l 0 5o 211 

10 0)0 

Since the motion is Simple Harmonic (why?), and remembering 

2tt 

that in S. H. M. displacement is 8 = 5 0 sin t, angular velocity is 


2 tv * 27r , . 2?r^ 

~7p do cos~ 7 p t, and maximum angular velocity is co 0 = 6 0 , 


equation 211 may be written 

NaHQ = Lq5 0 T . 

10 g 2tt 


From this we have 









THE BALLISTIC GALVANOMETER 


151 


Q = 


lOLoT # 
2tt aNF do 


212 


In terms of scale deflection 5 0 = , where r is the distance from 

2r 

scale to mirror. 


™ ~ 10 L 0 T 

Therefore Q = 


The current constant is 

10 L 0 


k = 


NaH2r 


(see eq. 183). 


T 

Hence Q = — kso = Q 0 s 0 213 

This ignores the damping due to air friction and induced currents. 
That is, So is the throw one would get if there were no damping. 

206. The Logarithmic Decrement. Equation 213 shows the 
relation between the quantity of electricity and the throw one 
would get if there were no damping. The correction to be applied 
to this expression is developed in the following discussion. 

When a galvanometer coil has acquired a momentum Ko> 0 as 
the effect of some momentary impulse, the motion will then be 
retarded and the coil ultimately brought to rest under the action 
of the elastic torsion of the suspension and frictional and magnetic 
damping. The equation of motion is 

— L 0 0 — c'co = Ka 214 

where K is the moment of inertia; a, the angular acceleration; 
L 0 , the moment of torsion of the suspension; 0, the angular 
displacement at any instant; go, the angular velocity at any 
instant; c', a constant, c'oo represents the damping torque 
due to air friction and induced currents, these both being very 
nearly proportional to the angular velocity. 

The motion differs from simple harmonic motion because of 
the term c' co. We have what is known as damped vibrations. 
Equation 214 may be written 

Kd 2 0/dt 2 + c' d0/dt + v Lo0 = 0 215 

The method of solving such a differential equation was sug¬ 
gested in the treatment of the equation of simple harmonic motion 
(para. 16). We will simply state that the solution in this case is 



where e is the Naperian log. base; T, the period of the motion; 
0o and c, constants. 





152 


ELECTRICITY 


We may verify this by differentiating twice with respect to 
time, we then get 

( 4 7T^ \ 

■JjY + c 2 y 0 = 0 217 

Eq. 215 in this form is 

d 2 0/dt 2 + ^ d ^+ |° 4 >= 0 218 

We thus see that 216 satisfies equation 215 or is a solution of it. 

It remains to interpret the constants, 0 O , and c. If the 
motion were S. H. M., equation 216 would be 

<t> = <j >0 sin ( Y t ) 

Hence we see that 0 O is the undampened amplitude or the value 
the initial throw would have had if undamped. If we find the 
ratio of the first throw, 0 1 , to the next throw, 0 2 , in the other di¬ 
rection we have 


0i 0 o e ~~ 4 219 

02 _3CX_ 

0 ce 4 

since the time, t, for the first throw to be attained is T /4 and 
for the next on the opposite side is 3T/4. This becomes 


01 cT /2 0] cT 

= e or log = 1 


The log (0i/0 2 ) is called the logarithmic decrement, X, and 
x = cT 
X 2 

Equation 216 thus becomes 


220 


221 


— 2 \ 
~T~ t 


(tP*) 


0 = 0oe 


sin 


222 


In the use of the ballistic galvanometer we are concerned with 
the first throw only. For this case equation 222 becomes, if we 
substitute the symbols s and s 0 for 0 and 0 O , 


s 0 e 


-X/2 


Therefore s 0 = s e ^ 

Equation 213 then becomes 



MAGNETIC INDUCTION IN IRON 153 

Q = Qoe X/ 2 s 223 

where s is the actual observed throw. 

If we expand e X//2 in powers of X and neglect higher powers 
than the first we have 

Q = Qo(l+ x/2)s 224 

Equation 224 is approximate and should be used only for 
small throws and for small values of X (not to exceed X = 1.0) 

X may be experimentally determined by simply observing 
two successive throws and taking the Naperian log. of their 
ratio. 

207. The Relation Between the Units of the Electromagnetic 
and Electrostatic Systems. It should be noted that by measuring 
the capacity of a plate condenser in electromagnetic units as 
outlined above and by computing the capacity in electrostatic 
units from the dimensions the relation between the units of two 
systems may be determined. 

MAGNETIC INDUCTION IN IRON 

208. Effect of Poles. If a bar of iron is placed in a magnetic 
field, H, Fig. 120, it becomes magnetized. The field at all points 
within and without the iron is modified by the presence of the 
free poles at the end of the bar. 



























154 


ELECTRICITY 


The resultant field is due to the superposition of the new field 
due to the free poles of the iron upon the original field. The 
resultant field will appear somewhat as shown in Fig. 121. If 
one considers a point within the iron the field is (H—F), where 
F is the field due to the free poles N and S. If the magnet is 
very long, the field at some point near the center of the bar, 
due to the free poles may be neglected, or if we have a closed 
ring in which case there are no free poles, then the field within the 
iron is merely H. 

The induction, B, within the iron, in the case of a short bar, is 
B = (H — F) + 4 ttA 225 

This is equation 164*. 

In the case of a very long rod, F, becomes negligible. 
In the case of a ring there are no free poles and F does not exist* 
In either case we have the expression for induction as 

B = H + 4ttA 226 

If we were to cut a narrow transverse gap across the iron free 
poles would appear on the faces of the gap. The induction as 
given in equation 226 is equal to the field in such a gap. 

209. Hysteresis. If a sample of iron is placed in a solenoid and 

a magnetic field 
applied step by 
step, we may 
study the rela¬ 
tion between this 
magnetizing field, 
H, and the vari¬ 
ous other quan¬ 
tities, viz., the 
intensity of mag- 
netization, A, 
the magnetic 
induction, B , 
and the permea- 
bility, ix. Fig. 
122 shows the 
relation between 
B and H as 
determined ex¬ 
perimentally for 
a sample of cast 
iron. The iron, 
originally unmagnetized shows an increasing magnetization, indi- 
catedbyB,asthe field, H, is increased from zero. The relation 

* Mo &hich occurs in eq. 164 need not be retained since its value is unity. 
Strictly, Mo is necessary to make the equation dimensionally correct. 



Fig. 122 








MAGNETIC INDUCTIOH IN IRON 


155 


between B and H is shown by oc (the virgin curve). At C 
the iron has become completely magnetized (said to be saturated ). 
A further increase in H, say AH, simply increases B by the 
amount AH. If when the iron is saturated we diminish the mag¬ 
netizing field to zero the curve, cd, is obtained. The induction, od, 
is due to the residual magnetism. Reversing the magnetizing field 
at this point and increasing it in this reverse direction gives the 
curve, def. The field, oe, necessary to reduce the magnetism to 
zero, is called the “coercive force.” At f the iron is saturated in 
the reverse direction. Diminishing the field, H, to zero, reversing 
and increasing the original direction gives us the curve, fgc. We 
have thus carried the iron through a complete magnetization 
cycle. We may note that the magnetization continually lags be¬ 
hind the magnetizing field. For instance, it does not become zero 
until sometime after the field has passed that value. This effect 
is called hysteresis and the curve cefgc is called the hysteresis loop. 
It will be shown later that the area within this loop is proportional 
to the energy consumed in carrying the iron through the magne¬ 
tization cycle. 

210. Experimental Methods for Obtaining the Hysteresis Loop 
for a Sample of Iron. (1) Rowland’s Method. The change in 
magnetic induction in a ring magnet is measured by observing 
the throws of a ballistic galvanometer. Two coils are wound on 
the ring, one used for producing the magnetizing field, the other 
connected to a ballistic galvanometer. The current in the mag¬ 
netizing coil is changed by properly chosen steps, and the throw 
of the galvanometer coil produced by the flux change through the 
second coil is read. The throw is proportional to the quantity of 
electricity flowing, which in turn is proportional to the change in 
flux. Since 

F _ Nd0 
10 8 dt 


and since E = R I we have Nd <f> = 10 8 RI dt 
or d (f> = 10 8 RdQ /N, and since d0 = adB 

we have the relation 


1T , 10 8 RdQ 
dB = —or Bi 
aN 

Using equation 224 this becomes 


10 8 RQi 

aN 


B, = 10 8 RQ 0 (1 + \ X) si /aN 227 

Thus the various changes in B may be computed from the 
corresponding throws of the ballistic galvanometer. The change 
in H, corresponding to a given change in B, is computed from 
the cnange m the current in the solenoid, since 
H = 4 ttNI/10/ 





156 


ELECTRICITY 


(2 s ) The Magnetometer Method. The iron in the form of a 
long rod is placed in a considerably longer solenoid. A magnetom¬ 
eter needle is placed at some point on the axis of the solenoid 
and the deflection observed as the magnetization in the iron is 
increased. A neutralizing coil, placed between the solenoid and 
the magnetometer, balances out the effect of the solenoid field. 
The field at the needle due to the magnetized poles in the iron is 
F = H e tan 5, (where H e is the horizontal component of the earth’s 
field. 5 the deflection of the magnetometer). Now 


(r -V (r+Z) 2 

where r is the distance from magnetometer to center of iron rod; 
/, half the length of the rod; and m, the pole strength. 

Also — = A 

a 

where A is the intensity of magnetization; a, the cross-section 
of the bar. 

4 ttNI 

Hence computing H from the expression H = — ■ - and using 

equation 226 we may obtain B for any value of H. 

211. The Work per Unit Volume to Magnetize a Sample of 
Iron. Let us consider either an anchor ring within a circular sole¬ 
noid or a long bar within a long straight solenoid. The following 
reasoning holds for either case. 

When the magnetization is increased the flux change, d0, intro- 


N d cf> 


I = ?? (eq. 178). Substi- 


duces an opposing e. m. f., E = - 1Q8( ^ . The work required to 

overcome this is dW = E Idt. ^ _ 

47rN 

tuting the value of E and I in the expression for dW we get 
™ 10 l N H d0 

47T 

which may be written as 

dW = fjdB. 

47r 


Integrating, we have 

W =fU HdB 

and work per unit volume is 

W =-LfHdB 
4:TT J 


228 


It will be noted that for a complete cycle j HdB is the area of 
the hysteresis loop. 








MAGNETIC INDUCTION IN IRON 


157 


212. Relation between Flux of Induction and Ampere Turns. 

The work done in carrying a unit pole once around a path encircling 

N turns of wire carrying current I is 27rrX^~ (see eq. 177) or 


47rNI 
10 ' 


229 


230 


Work per unit pole is also given by the expression 
w = jHdl = Hj/i + H,Z a + • • • 

Therefore NI = -^-(H,/,+ H 2 / 2 + • • •) 

47r 

Hi, H 2 • * * represent the magnetizing field at the various 

points of the path. This may be written 

NI = +A-+ •••.)* 231 

47r ai/ii a 2 /x 2 

213. The Magnetic Circuit; Analogy with the Electric Circuit. 

Work per unit pole is given by the expression 

dw = Hd/ =- - C ^ or 


w = </>(- 


h 


SL/JL 


+ 


232 


aijui a 2 ju 2 

For a circuit of constant cross section and permeability, 

w =— $ 233 

SL/JL 

The analogy is apparent between this expression and Ohm’s 
law for a complete electric circuit, viz., 

E =—I 
ac 


where c is the conductivity; the entire resistance of the circuit; 

and E, the work to carry unit quantity of electricity completely 
around the circuit. w, by analogy, is called the magnetomotive 
force (m. m. f.); I /sl/jl, the magnetic resistance, is named 
the reluctance. The c.g.s.unit of m. m. f. is the erg per unit pole. 
Usually in practice magnetomotive force is expressed in ampere- 
turns. Ampere-turns are obtained by multiplying Eq. 233 by 
10/47T (thus giving Eq. 231). For a portion of an electric circuit 
the potential difference is expressed by 


*In the case of a bunched windings instead of uniformly distributed sole¬ 
noid as above assumed, the theory given is only an approximate treatment, 
since in this case the magnetizing field (in the iron) is not the same for all 
portions of the circuit and therefore the intensity of magnetization is not 
uniform. This means free poles and magnetic leakage and consequent vari¬ 
ation in 0. Eq. 231 however gives approximate values even in such cases. 







158 


ELECTRICITY 


Pd, =—-I, and E = Pdi + Pd 2 + • • • • 

&lCi 

For a portion of a magnetic circuit the magnetic potential differ¬ 
ence (also called m. m. f.) is 

(m. m. f.)i = —^-0, and m. m. f. = (m. m. f.)i + (m. m. f .)2 + * * 

aijui 

Or m.m.f. = 0 ( + —— + * * * *) 234 

ai/ii a 2 /x 2 

Just as e. m. f. = i (—-—1--—b * ' *) 

aiCi a 2 c 2 

Each term on the right of 234 being the m. m. f. for the particular 
part of the circuit. 

Expressed in ampere-turns equation 234 becomes 

M.M.F. =-!%(——|——+ • • •••)* 235 

47T 2 LiHi a 2 jU 2 
Or we may use the form 

M, M. F. = Nil! + N 2 I 2 + • • • • 236 

The ampere-turns, NJi, N 2 I 2 , for any value of flux density 
for the various forms of iron may be obtained from published 
curves such as those of Fig. 123. 

214. Method of Solving Problems on the Magnetic Circuit. 
In solving practical problems equation 236 is convenient since 

for any kind of iron the ampere 
turns per cm. for any desired 
flux density can be obtained 
from curves for the iron in 
question. Such curves for cast 
iron, wrought iron and cast steel 
are shown below. The following 
typical case will show how 
problems on the magnetic circuit 
are handled. Suppose a mag¬ 
netic flux of 1,500,000 lines is 
desired across the air gap of a 
dynamo and suppose the field 
magnet to be of wrought iron and 



Fig. 123 


* It should be noted in the application of eq. 235 that /* apparently 
•depends upon the magnetic circuit as well as upon H and the particular 
specimen of iron. This is not really so, as the following case will show. Con¬ 
sider two rings of the same size and same kind of iron, each wound with the 
same number of turns and let one ring have a narrow gap sawed through it. 
With the same current in each coil the magnetizing field is not the same. 
For the ring with the gap the magnetizing field is (H - F) where F is the 
field due to the free poles exposed at the gap and H, the field due to cur¬ 
rent in the solenoid. For the other ring the magnetizing field is H. The 
relation B =/j.K holds for all cases if we remember that H stands for the 
actual magnetizing field. 













-FLaX PErtSlTY 


MAGNETIC INDUCTION IN IRON 


159 


the armature core of cast iron. Let the wrought iron portion 

have a cross-section of 100 sq. cm. and length of 50 cm., the cast 

iron portion a cross-section of 200 sq. cm. and a length of 20 cm., 

and the air gaps a cross-section of 200 sq. cm. and a total length of 

1 cm. How many ampere turns are needed? 

Solution: First find the flux density or induction for each 

part of the magnetic circuit. Thus for the wrought iron portion, 

<f> 1,500,000 •C' 4-t, 

Bi = — = ~— = 15,000. For the cast iron portion, 

ai 100 

B 2 = — = = 7,500. We find from the curve that for 

a 2 zUu 


wrought iron the number of ampere turns per cm. (corresponding 
to Bi = 15,000) is 20 and hence Nib = 20 x 50 = 1,000; for 
the cast iron, the number of ampere turns per cm. (corresponding 
to B 2 = 7,500) is 50, and N 2 I 2 = 50 x 20 = 1,000. For the 


air gap 


B a 


a 3 


1,500,000 
200 


= 7,500. 


For air 






































































160 


ELECTRICITY 


A nr- TSj T 

B = H and H - ^ or N 3 I 3 = 10 HI (eq. 230). 

Thus N 3 I 3 = L* x 7500 x 1 = 5,950. Hence the 
47r 

total number of ampere turns required is NJi + N 2 I 2 + N 3 I 3 
= 7,950. 

ELECTROLYSIS 

215. Conduction of Electricity. Theory. In the case of metals 
the movement of small negative charges (electrons) constitutes 
what we call the flow of electricity. These electrons are constit¬ 
uent parts of the atoms. Indeed we believe the atom to be made 
up of configurations of negative electrons in rapid motion about a 
positive nucleus. Under the influence of an electric field there 
ensues a drift of these negative electrons up the field. This 
movement we know as an electric current. 

In the case of electrolytes the flow of electricity is explained 
in another manner. In any conducting solution such as a salt 
dissolved in water, some of the molecules are found to break up 
into positive and negative portions called ions. For instance, 
in a metallic salt, the metal atom with its + charge separates 
from the negative acid radical, the effect of the solution being to 
produce a weakening of the electric forces holding the molecules 
together. This separation is known as ionization and the parts 
are called ions. There are thus both positive and negative 
ions, both are free to move, and, according to kinetic theory, 
they normally are in a state of rapid random motion like the mole¬ 
cules of a gas. The solution of course contains in addition the 
molecules of the solvent and undecomposed molecules of the 
solute. All are in the same state of rapid random motion. When 
an electric field is applied to opposite boundaries of such a solution 
a definite drift of the positive and negative ions naturally ensues, 
the positive (metal ions) being attracted to the negative electrode, 
the acid radical to the positive. The ions upon giving up their 
charge will either react chemically with the solution or with the 
electrode or will remain attached to the electrode or will be 
liberated as a gas. 

216. Faraday’s Law of Electrolysis. Faraday announced the 
following laws as the result of experiment: 

(1) The mass of substance liberated at an electrode is pro¬ 
portional to the current and the time or to the quantity of electric¬ 
ity passing. Thus 

m = elt 237 

e is a proportionality constant. It is called the electro-chemical 



ELECTROLYSIS 


161 


equivalent and may be defined as the mass of the substance 
liberated by one ampere in one second. 

( 2 ) The mass of a substance liberated by a given current in a 
given time is proportional to the chemical combining weight of the 
substance that is, the atomic weight, a, divided by the valence, 
v. Thus 

e oc a/v or e = h (a/v) 238 

where h is the proportionality constant, h is the electro¬ 
chemical equivalent of hydrogen. 


PROBLEMS 


1. If e, Equation 237, has the value for copper 0.000329, 
for hydrogen 0.00001044, for silver 0.001118, zinc 0.000338, 
find the number of coulombs required to liberate a gram equiva¬ 
lent of each metal. How would you expect the values to com¬ 
pare? 

2 . Computations based on kinetic theory show the number of 
atoms per gram of hydrogen to be approximately 6 x 10 23 . Each 
hydrogen atom or ion carries an elementary + charge (which 
probably means it is short one negative electron required for 
neutral condition). The quantity of electricity required to 
liberate a gram equivalent of hydrogen is 96550 coulombs. Find 
the size of the elementary charge, (the electron) in terms of the 
coulomb. 

3. Knowing the electro-chemical equivalent of silver show 
how by electrolytic methods an ammeter may be calibrated. 






217. Energy Equation of an Electrolytic Cell. In case current is 

sent through an electrolytic cell, 
such as one consisting of two 
platinum electrodes and a dilute 
solution of H 2 SO 4 , there will 
occur in addition to the evolu¬ 
tion of hydrogen and oxygen 
at the plates an accumulation 
on or absorption by the plates 
of these gases, hydrogen at the 
cathode, oxygen at the anode. 
On removing the applied e.m.f. 
Flg ' 125 and connecting the electrodes 

externally through a galvanometer the existence of a potential 
difference between the plates will be indicated by a flow of current. 



Before the e. m. f. applied to the electrolytic cell has been removed there 
occurs an accumulation on the kathode of H molecules which have given up 
their (+) charges, these having been neutralized by the negative drift in the 
external circuit in the other direction. On the anode occurs also a neutral 
accumulation of oxygen, O. The “solution tension” (see paragraph 219) 













162 


ELECTRICITY 


between the solution and the hydrogen layer being greater than that between 
the oxygen and the solution a back e. m. f. is developed, the existence of which 
is shown on removing the applied e. m. f. and closing the external circuit. 
We get through the external circuit a current from anode to kathode and 
through the cell from kathode to anode. This back e. m. f. is called a polar¬ 
ization e. m. f. 

The current in the external circuit will be from the anode to the 
cathode and that through the cell from cathode to anode. In 
other words a reverse e.m.f., Ei, is thus shown to have been 
established. This e.m.f. is called a polarization e.m.f. and the 
plates are said to have become polarized. Therefore in sending 
current through an electrolytic cell it is necessary to supply 
energy at the rate Eil to overcome this polarization e.m.f. The 
energy required to overcome the ohmic resistance of the cell is 
RI 2 . Finally we must supply energy to produce the chemical 
reactions which occur at the electrodes. In the cell under con¬ 
sideration the reaction occurs at the anode, the S0 4 ion giving up 
its charge to the electrode and thereupon decomposing to S0 3 and 
O, the oxygen atom combining with another and rising as gas, 0 2 . 
The S0 3 attacks the water producing H 2 S0 4 which ionizes into H, 
H and S0 4 , the hydrogen ions then migrating across the cell. 
The amount of water decomposed is by Faraday's laws propor¬ 
tional to the current, hence the energy to be supplied per second 
for this purpose is proportional to the current and may be ex¬ 
pressed as E 2 I. E 2 is called the counter e.m.f. of the cell. Thus 
the energy equation for an electrolytic cell is 

Pd I = RI 2 + Eil + E 2 I 239 

or Pd + RI + Ei -f- E 2 240 

We may obtain relative values of E] and E 2 for the simple cell 
shown above as follows: The decomposition of one gram of 
water requires 162 x 10 9 ergs of energy. One c.g.s. unit of 
current decomposes 0.000933 grams of water per second. Hence 
the energy to be supplied per c.g.s. unit of current per second is 
148 x 10 6 ergs which is 1.48 volts. This is the E 2 of equation 
240. The minimum e.m.f. necessary to get any current through 
such a cell is about 1.67 volts. Hence the value of Ei is about 
0.2 volts. 

218. The Voltaic Cell. In certain types of electrolytic cells the 
chemical reactions serve as a source of energy. Thus when zinc 
is substituted for one of the platinum terminals in the cell above, 
the reaction between S0 4 ions and zinc supplies more energy than 
the decomposition of H 2 S0 4 requires. 

If the terminals of such a cell are connected, a steady current 
will flow, the attraction of the zinc for the S0 4 ions continually 
maintaining a supply of negative electrons on the zinc while 
across the cell the H, H ions are supplying positive in equal pro- 


ELECTROLYSIS 163 

portion to the kathode. The action will continue until all the 
H 2 S0 4 has become ZnS0 4 when all action will cease. Such a 
cell in which chemical reactions at the electrodes supply the 
energy necessary for the passage of current is called a voltaic 

cell. 

219. Theory of the Chemical Action of a Simple Voltaic Cell. 

The following approximate explanation will serve to give an idea 

of what takes place within 
the cell. The chemical 
force (attraction) pulling 
the metal ions into the 
solution is known as solu¬ 
tion tension. When ions 
leave the Zn or Cu and go 
into the solution with their 
+ charges they leave the 
metal negative. There is a 
certain definite solution 
tension for each metal in a 
given solvent. For the case in question Zn ions will leave the 
zinc until a certain definite potential difference is established 
between the Zn and the solution. This pd indicates a force pulling 
the ions back to the metal. At a certain definite pd, equilibrium is 
established. The tension of H2SO4 for Zn is greater than for 
Cu. Hence the Zn will be reduced to a lower potential than 
the Cu, or the Cu will be positive to Zn. 

If the Cu is connected externally to the Zn then current will 
flow from Cu to Zn. Thereupon the pd. between the Zn and 
the solution is reduced or the equilibrium disturbed and Zn ions 
go off into the solution tending to restore the balance. Like¬ 
wise at the copper terminal the equilibrium is disturbed, the 
potential of the Cu being further lowered by the passing of (—) 
around the wire to the Cu. Hence to establish the equilibrium 
again Cu or H ions return to the metal. Suppose we have reached 
a stage where all the Cu has returned to the metal, then for every 
Zn ion that goes into solution an H pair goes out at the Cu terminal. 
When all H ions are gone then the action stops for the solution 
is entirely ZnS0 4 . 

220. Energy Equation of a Voltaic Cell. The equation is that 
already developed for an electrolytic cell. There Lave been no 
new factors introduced. There is however no externally applied 
pd. The left hand member of the equation is therefore zero. 
Also the sign of E 2 I is changed since there is a net energy supply 
rather than consumption. Hence the energy equation is 

E 2 I = RI 2 + Ejl 
E 2 — Ei = RI 



whence 


241 












164 


ELECTRICITY 


(E 2 — Ei) is the e.m.f. of the cell. E 2 I is the energy supplied 
per second by the chemical reactions and may be measured as 
heat if the reactions are carried out in a calorimeter. 

There are many types of voltaic cells. Usually the anode is 
of zinc, the cathode of any inert material which is a good electrical 
conductor. The electrolyte is decomposed by the current, the 
negative ion (anion) reacting with the anode, the positive ion 
(cation) appearing at the inert cathode. If the cation happens to be 
hydrogen and is allowed to collect the polarization e.m.f. above 
mentioned is built up and will reduce and may stop the action of 
the cell. The available energy and therefore the efficiency depends 
upon the removal of the hydrogen. This is usually accomplished 
by supplying oxygen in some form at the cathode. (The larger 
treatises on Physics should be consulted for a description of 
various commercial voltaic cells.) 

221. Source of Electrical Energy in a Daniell Cell. The re¬ 
actions to be considered are simply the formation of ZnS0 4 and the 
decomposition of CuS0 4 . When the electro-chemical equivalent 
of zinc goes into solution at the anode, the electro chemical equiva¬ 
lent of copper must go out of solution at the cathode this being 
required for the transfer of one coulomb of electricity across 
the cell. When zinc reacts with H 2 S0 4 there are two processes 
involved, viz., the formation of ZnS0 4 , resulting in the generation 
of the heat formation , and the decomposition of H 2 S0 4 with 
liberation of hydrogen, requiring the absorption of the heat 
of decomposition of H 2 S0 4 . The processes taking*, place in the 
reaction between hydrogen and CuS0 4 are the decomposition 
of CuS0 4 with liberation of Cu. requiring the absorption of 
the heat of decomposition and the formation of H 2 S0 4 requiring 
the generation of the heat of formation of H 2 S0 4 . The net effect 
is the liberation of energy representing the difference between 
the heat of formation of ZnS0 4 and that of decomposition of 
CuS0 4 under the given circumstances. 

The heat of formation of the electro-chemical equivalent of zinc 
sulphate determined experimentally for the conditions obtaining 
in a Daniell cell (that is of 32.44 x 1.044 x 10* 5 grams) is 1.304 
calories (heat generated). 

The heat of decomposition, determined experimentally under 
the same conditions, of the electro-chemical equivalent of copper 
(that'is of 31.8 x 1.044 x 10 -5 grams) is 1.042 calories (heat ab¬ 
sorbed) . 

The electrical work done in transferring one coulomb is EQ 
joules or EQ x 10 7 ergs, where E is the e.m.f. of the cell and Q is 
one coulomb. Equating this product to the heat liberated 
(0.262 cal. or 1.09 x 10 7 ergs), 


THERMOELECTRICITY 


165 


E Q 10 7 = 1.09 x 10 7 ergs 
Or E = 1.09 volts 

This agrees with the observed value. 

222. Secondary (or Storage) Cells. A cell in which the chemi¬ 
cal reactions may be reversed by causing current to flow through 
the cell in the reverse direction and which thereby may be restored 
to its original condition constitutes a storage cell. The products 
resulting from the chemical action when the cell is producing cur¬ 
rent must in such a cell be conserved in the electrolyte. The fol¬ 
lowing simple cell will illustrate the principle. If current is passed 
through a cell composed of two lead plates placed in a weak solu¬ 
tion of H 2 S0 4 , the oxygen liberated at the (+) electrode reacts 
with the lead forming Pb0 2 . When the charging current is re¬ 
moved the cell will be found to act as a primary battery. 

The terminals will still be of the same sign but current will flow 
through the cell from the negative electrode to the positive and 
around the external circuit from positive to negative. This is now 
the behavior of an ordinary primary cell. It will be noted that 
the direction of the flow of ions on discharge is reversed from their 
action on charging, the hydrogen ions, H, with their positive 
charges migrating to the oxide coated plate, reducing the oxide 
to lead and forming water. The cell will give current until reduc¬ 
tion is complete. The actual lead storage cell is not as simple in 
its construction as this and the reactions are more complicated. 
The simple cell merely serves to illustrate the principle. 


PROBLEM 


If we have a number of cells (all alike) and wish to get the 
maximum current possible through a given external resistance, 
what combination of series and parallel arrangements should we 
adopt ? 

Suggestion: Let p = number of series, q = number of cells in 
each series, then . 

n (total number of cells) = pq, and p = n/q 

Then 


I = 


qE 

R + qr 


P 


where R = the external resistance, r = the resistance of each cell. 
Apply maximum and minimum methods, varying q until I 
becomes a maximum and thus determine the conditions sought. 


THERMO-ELECTRICITY 

223. Thermo-Junctions. If two dissimilar wires are joined to¬ 
gether forming a closed circuit and one junction heated a current is 



166 


ELECTRICITY 


found to flow in the circuit. The e.m.f. producing such a current 
is known as a thermal e.m.f. Experiment shows that such an 
e.m.f. depends solely upon the material of the wires, the mean 
temperature, and the temperature difference of the junctions. 

For a given junction the relation can usually be expressed by 
the empirical equation 

E = a + bT + cT 2 242 

By differentiating this with respect to temperature we have 

dE/dT = b+2cT 243 

E is the thermal e.m.f.; T, the temperature of the hot junction. 
dE/dT is called the thermo-electric power, a, b and c are 
constants; c is the slope of the thermo-electric power, tempera¬ 
tures line and depends only upon the materials of the wires; b 
is the slope of the e.m.f. temperature line at the point, T = 0, and 
depends only on the material of the wires; a is the thermal 
e.m.f. when the hot junction is at zero and depends upon the 
material of the wires and the temperature of the cold junction. 

If the cold junction is kept at zero degrees, a = 0 and equation 
242 becomes 

E = bT + cT 2 244 

224. Applications of Thermo-Junctions. The thermo-junction 
furnishes us with a very valuable instrument for the measurement 
of temperatures. By the use of suitable electrical measuring 
devices such as galvanometers and potentiometers, temperature 
measurements of extreme refinement may be made and over very 
large ranges. With a junction of platinum and an alloy of plati¬ 
num and 10 percent rhodium temperatures may be measured 
from liquid air to nearly the platinum melting point (1753°C) or 
over a range of some 1800° C. A temperature difference of 0.1 
degree may easily be measured. 

ALTERNATING CURRENTS 

225. The Fundamental Equations of the Alternating Current 
Circuit. We have seen, § 197, how an alternating e. m. f. is 
produced. A knowledge of the characteristics of the current 
which results from impressing such an e. m. f. on a circuit, and of 
the relations between current, e. m. f. inductance, capacity and 
resistance are of very great importance to the engineer. Probably 
the greater part of applied electricity lies in the field of alternating 
currents. Therefore a brief survey of the fundamental relations 
which form the basis of alternating current theory will be under 
taken in this section. 

We have already called attention to the phenomena of self 
and mutual induction which come into play whenever the current 
in a circuit is undergoing change. In the case of alternating 


ALTERNATING CURRENTS 


167 


current, the current is continually changing or alternating from 
positive to negative maxima. Equation 199 applied to this 
case, assuming that the circuit contains resistance and inductance 
only, becomes 

J T 

E 0 sin cot = L tt + R I 245 

at 

From the energy standpoint we might have written the equation, 

E 0 I sin cot = L I + R I 2 , where E 0 I sincot expresses the 

rate at which energy is being supplied at a given instant; LI dl /dt 
the rate at which energy is being stored in the magnetic field; 
and R I 2 , the rate at which energy is being converted to heat. 
This leads to the equation above. 

In case the circuit contains a condenser which must be charged 

the energy equation is E 0 1 sin cot = L I ^ + E C I + R I 2 , where 

E c is the voltage across the condenser at the given instant, E G I 
then being the rate at which energy is being stored in the con¬ 
denser. The equation states that the energy supplied at rate 
E I sin cot is being converted into electromagnetic energy (energy 
of the magnetic field) at the rate LI dl /dt, into the electrostatic 
energy of the charged condenser at the rate E C I, and into heat 
at the rate R I 2 . Cancelling the I which occurs in every term, 

and substituting ^for E c (see § 139) we have 

E 0 sin cot = L + g + R I. 246 

This is the general fundamental equation of alternating current 
theory. It leads to eq. 245 in case the capacity is infinite which 
means a closed circuit with no condenser. 

226. Circuit with Inductance and Resistance. The Equation 
for Current. The equation 

E 0 sin cot = L-^-+ R I 

is a simple differential equation with current and time the varia¬ 
bles. The general solution of this equation or the function of 
I which satisfies it under all conditions is 

I = A sin cot + B cos cot 247 

A and B are constants whose values can be obtained by substi¬ 
tuting this expression in the original and the expression for dl /dt 
obtained by differentiation. [The student may satisfy himself 
that this equation is the solution of the original by substituting 
in the original after the proper values of A and B have been 
determined]. 


168 


ELECTRICITY 


We proceed to the determination of A and B as follows, 

^ 7 - = Ac 0 cos cot — Bco sin cot 
dt 


24g 


Substituting this and expression 247 in the original we get 
E 0 sin cot = Leo A cos cot — Leo B sin cot 

+ RA sin cot + RB cos cot 

We may now evaluate A and B by assuming arbitrary convenient 
values for t. 

Thus, when t = 0, we have 0 = LcoA + RB. When t = ^,we 

2 co 


have E 0 = — LcoB + RA . 

From these two equations we get, A = 


RE C 


,andB = . 


—LcoE c 


L 2 co 2 + R 2 L 2 co 2 + R 2 ‘ 

Substituting these in the general solution for I (eq. 247), we get 
E 

I = J^ 2 qj 2 _|_ ( R S * n - Leo cos cot) 

R 

If we write^= ^_j _.^ 2 = cos 0, it will follow that 

V W+R’*™* 

where 0 is merely any angle whose cosine has the numerical 
R 

vralue ^/ j ^ 2co 2 Asking these changes we have 

F 

I = — 0 — (sin cot cos 0 — cos cot sin 0) 

VLV + R 2 

This may be written in the form 
E 0 


I = 


Or 


VL 2 co 2 + R 2 
I = I 0 sin (cot — 0 ) 


sin (cot — 0 ) 


249 


This equation shows that the current lags behind the electro¬ 
motive force by an angle 0 , and that the maximum value of the 
E 

current is I 0 = , 0 -- We may note also that the angle 

\/L 2 co 2 + R 2 . 

of lag, 0 , is given by the equation 
6 = tan ' 1 (Lco/R) 


250 


The quantity, yTV + R 2 is called the impedance , Z, of the 
circuit. The quantity, L co, is called the reactance , X, of the 
circuit. If the inductance, L, is zero, the impedance becomes 



















ALTERNATING CURRENTS 


169 


a pure resistance and the angle of lag becomes zero. If the 
resistance is zero the impedance becomes a pure reactance and 
the angle lag becomes 90°. 

227. Vector Representation of Voltages in an A. C. Circuit. 

We have seen in the dis¬ 

cussion on simple harmonic 
motion that a quantity which 
varies as a sine or cosine 
function of the time may be 
represented graphically as the 
projection on a diameter of 
a radius vector which is 
rotating with uniform velocity. 
The quantities, E 0 sin cot, 

L d I /dt, and RI, are such quan¬ 
tities. E 0 sin cot may be 
represented as the projection 
on th3 y-axis, Fig. 127, of the 
revolving vector, E 0 . The 

quantity, R I, which equals RI 0 sin (cot—0) may be represented as 
the projection on the y-axis of the revolving vector, R I 0 , which 
rotates with the same velocity, co, as the vector, E 0 , but lags 

behind it by the angle, 0. The quantity, L dl /dt, which equals 

LI 0 co cos (cot - 0) or L I 0 co sin (cot —0+ 7r/2), may be represented 
as the projection of the revolving vector L I 0 co which is 90° 
ahead of the vector R I 0 in phase. The vector quantities R I 0 
and LcoTo are thus merely the rectangular components of the 
vector E 0 . 

Current may be represented on the same diagram as the pro¬ 
jection on the y-axis of the radius vector I 0 which is in phase 
with the vector R Io. Its value as well as that of the applied 
e. m. f. at any given instant may be obtained from the diagram. 
If instantaneous values of applied e. m. f. and current are plotted 
against time as abscissa we get curves such as are shown on Fig. 
128. 











170 


ELECTRICITY 


228. Circuit with Inductance, Resistance, and Capacity. As 

we have seen the equation for the circuit in this case is 

E 0 sin cot = L J gt+-§-+ RI 

dO dl d 2 0 

Now I = and-g^- = -g^-, hence the equation may be written 


E 0 sin*t = L^g-+R-g. +§-, 251 

the variables now being Q and t. The solution of this equation is 
similar to that of eq. 245, namely 

Q = A sin cot + B cos cot 

If this equation is handled in an exactly similar manner we 
finally obtain for current 

E 

1 = V( l/Cw-Lo»)*+ R 4 Sm( " t + d> 252 

Or I = Io sin (cot + 6) 


The impedance here is Z = V( 1/Cco— Lco) 2 +R 2 , 253 


, „ . ' l (l /Cco — Leo) 

and 0 = tan ——g-- 254 

This equation reduces to eq. 245 when C = oo, that is, for a 
closed circuit containing no condenser. 

It will be noted that if 1 /Cco > Leo, the angle 0 is positive or the 
current is ahead of the electromotive force in phase. If inductance 
and resistance are negligible the current will lead by 90°. By 
adjusting capacity and inductance the phase of the current 
relative to the e. m. f. may be shifted at will. If the values of 
C and L are so chosen that 1 /Cco = Leo, the impedance is a 
minimum and therefore I 0 , a maximum. In this case the phase 

2 7 r 

angle 0 is zero. Since co = —, where T is the period of the 

cycle, we have for this_state of affairs 

T = 2tjVLC 255 

or since co = 27m, where n is the frequency, we may write 
this as 


1 

27r\/LC 


256 


The particular frequency which satisfies this relation is called 
the natural frequency of the circuit. These considerations have 
important application in the field of oscillating circuits and radio 
telegraphy. 









ALTERNATING CURRENT THEORY 


171 


229 Vector Diagram for Circuit with Capacity, Inductance 
and Resistance. The revolving vector, E 0 , in this case is resolved 
into three components, R I 0 , which by eq. 252 leads the vector, 
E 0 , by angle, 0, which may be intrinsically positive or negative 
depending upon the relative magnitudes of the quantities which 
determine it, (see eq. 244;) Lcol, 0 which is 90° ahead of the vector, 
RI 0 ; and I 0 /Cco which is 90° behind the vector RI 0 . Lcol 0 
is the maximum value of Ldl 0 /dt. It is obtained by multiplying 
equation 252 by Land differentiating which gives 
L dl 0 /dt = Leo I 0 sin(cot+ 0+ 7 t/ 2). The radius vector, Leo I 0 , accord¬ 
ing to this equation leads the vector, RI 0 , by 90°. The vector,-^, 

is obtained by taking the maximum value of g' 

Now g = gj~I dt =gj*l b sin (cot + 0) dt or 

§" = ~A cos (wt + 9) = A sin ( -“ t + d - J r 

Thus this vector is 90° behind the vector, RI 0 . 

The projection of these 
three vectors on the y-axis 
at any instant gives the re¬ 
sultant e. m. f. 

It will be convenient for 
future reference to write 
equation 246 in the follow¬ 
ing form, using the rela¬ 
tions above, E 0 sin cot 
= Lcol 0 sin (cot + 0 + 7 t/2) 

+ -i-sin (cot+ 0 —7r/2) 

+ R I 0 sin (cot + 0) 257 

230. Average Current and 

E. M. F. The instantaneous 
value of the e. m. f.isE 0 sin cot. 
The average value of this for a 



“if 


E 0 sin cot dt. This gives average e. m. f. 
=o 

T " - - 

average e. m. f. for a half cycle (beginning at point E = 0) 


complete cycle is 

• / L _ 

for whole cycle = — [cos «tj ( _= °- smce T 


2ir 

co 







172 


ELECTRICITY 


__=• r 

T to L 


cos cot 


-|t=T/2 ( E( , 

J t - ° 7r 

Likewise for current, the instantaneous value is Io sin (cot + 6) 

j f*<i)t-\-0 = ir 

Average current for whole cycle = ^ 


U 


L 0 sin 

ut+0=O 


258 


(cot-f-0) dt 


Jo_ 

Tco 


r "1 wi 

cos (cot + 6) 

L J « 


«t+0 =27T 


t-H=o 


= 0 


Average current for a half cycle (beginning at point I = ,0) is 


in* 

T JV. 


{(*t+0)=* J r- 

sin (cot + 0) dt = cos 


-1 6>t + 0 = 7T 


Io 


cot —0 — 

Jwt + 0=O T 


259 


=0 

231. Virtual Current and E. M. F. In all alternating current 
work ammeters and voltmeters are used whose action depends 
upon the square of the current or e. m. f. The readings of these 

instruments hence are 
independent of the sign 
or direction of the cur¬ 
rent or e. m. f. The 
reading of a hot wire 
ammeter, for instance, 
depend upon the heating 
effect. Fig. 130 will 
illustrate the principle 
of operation here. As 
the wire ab which carries 
the current elongates 
with rise of temperature 
Readings 





a 

Fig. 130 

the spring will cause the pointer to move over the scale, 
are proportional to R I 2 . Another instrument consists of a small 
coil mounted at the center of a large coil and with its plane held 
by a spring at right angles to that of the large one. If the two 
coils are connected in series the angle of twist of the small coil 
and therefore the deflection of a pointer attached to it will be 
proportional to the square of the current. This the student may 
show. "With alternating current the deflection of these instru¬ 
ments will be proportional to the average square of the current 
or voltage and they will normally be calibrated to read the square 
root of the average square current or voltage. This is called the 
virtual (or ammeter or voltmeter) current or voltage. 

The relation between the virtual current or voltage and the 
maximum current or voltage is of considerable importance. 
We have for the square of the instantaneous voltage 
E 2 = E 0 2 sin 2 cot 



ALTERNATING CURRENT THEORY 


173 


For the average value of E 2 for the whole cycle or during the 
time ,T, we have 


mean square e. m 


• f. = y (e,, 2 si 
J t= 0 


E 0 2 

/*t=T 

dt 

1 t=0 

W 0 C_j <N 

W S w 

ii 

2T 

2T 

Eo 2 

Eo 2 

t 

sin 2cot 

2 

2coT 


sin 2 cotdt 

=T 

(1—cos2cot) dt 

t =0 
*t=T 

cos 2cot dt 
t=o 

2tt 

t =T -- 


The second term is zero hence the root mean square or virtual 
e. m. f. is given by 

E v 


V2 


260 


In exactly similar way it may be shown that the root mean 
square or virtual current is given by 


i - ^ 
Iv ~y/2 


261 


232. Power in an Alternating Current Circuit; Power Factor. 

The value of the power at any instant in any circuit is E I. 
In an a. c. circuit since both E and I vary we will be interested in 
the average power for a cycle. This will be given by the ex¬ 
pression 


Average power 


le ins 

J t=o 
= sin cot 

E 0 io p: T 

— I sir 

J t =o 


Since E and I are the instantaneous values we may write this 

i r =T 

Average power = | E 0 sin cot X Io sin (cot — 0) dt 262 

Since sin (cot — 0) = sin cot cos 0 + cos cot sin 0 we have 

= T 


Average power 

This may be written 

. E o locos0 

Average power = 


sin cot (sin cot cos 0 + cos cot sin 0) dt 
=o 

f. 


sin 2 cotdt 

J t =o 


E o I o sin0 


in0 f. 1 
— I sini 

J t =o 


= T 

sin cot coscot dt 
263 


Remembering that sin 2 cot = J(1—cos 2cot) 

and sin cot cos cot = §sin2cot, we have 






174 


ELECTRICITY 


Average power = 


Eolocosflj^ 


2T 


(1 —cos2cot)dt 


Ei 


T 

sin2cotdt 


E o I o cos0 
2T 


r* n 

J t =0 J t 


T 

cos2cotdt 

J t =o 

t=T=- 


^fsin' 

J t =0 
ft= T 

I sin2 

J t =o 


E o losin0 


t=T 

sin2cotdt 


27r 


JE o I o cos0 — E ° 2 ^ S ^ [^ sin2cot ] + |^£cos2cotj 

t=0 


t=T = 


t=0 


27T 


The last two terms are each zero, therefore 

average power = iE o I o cos0 This may be written 
average power = E v I v cos0 264 

The fact is here shown that the product of voltmeter and ammeter 
voltage and current does not give power unless the lag angle, 0, 
is zero. Cos 0 is called the power factor of the circuit. 

233. Equation 264 may be interpreted in terms of the vector 

diagram. The vector, 
I 0 , represents the maxi¬ 
mum value of the current. 
It lags behind the vector, 
E 0 , by the angle, 0, 
assuming these vectors 
are rotating counter¬ 
clockwise with velocity, 
co. We may resolve the 
vector, I 0 , into a com¬ 
ponent, I 0 cos 0, in phase 
with E 0 and a com¬ 
ponent, I 0 sin 0, 90° 
behind E 0 . The in¬ 
stantaneous power, con¬ 
sidering the in-p h a s e 
component is E 0 sin cot 
X I 0 cos 9 sin cot or 
E 0 J 0 sin 2 cot cos 9. 



The average power is 
expression 264. The 


E o I o cos0 


T , 

I sm 2 
J t =o 


T 

instantaneous 


cotdt. 


This leads to 


power considering the 


90° out-phase component is E 0 sin cot X I 0 sin 9 sin (cot —— 


or 













ALTERNATING CURRENT THEORY 


175 


E 0 I 0 sin 6 sin cot cos cot and the average power is 


E, 


J t=0 


cot cos cot dt. 


This is identical with the second term of eq. 263 which was 
shown to be zero. This component is called the wattless current 
since the consumption of power by it is zero. 

If the resistance of the circuit is zero, 6 is 90° (eq. 250, eq. 
254) and the average power is zero (eq. 264), or the current 
is entirely wattless. The physical meaning of this is that during 
the part of the cycle in which the current is increasing, energy 
is being stored in the magnetic field, and during the part in which 
current is decreasing this energy is restored to the circuit. 

234. The Wattmeter. This instrument consists essentially 

of a small low 
resistance coil, 
ci, connected 
directly into 
the circuit and 
suspended at 
the center of 
a larger coil, 

c 3 , so that 
the planes of 
the two coils 
are at right 
Fig. 132 angles. The 

coil, c», is connected in series with a large resistance, R, and 
shunted across the line. The torque between these two 

coils is proportional to the product of their respective currents. 
Thus torque = k, I I'. 

I = I„ sin («t-0) and I' = sin (<ot -6'). R is 

V L 2 co 2 ~T R 

chosen so large as compared with the reactance term that the 
latter may be neglected and therefore 0' (which is tarn 1 Leo /R) may 
be considered zero. We have then 

k I E 

Instantaneous torque = —= 5—- 0 sin (cot — 0) sin cot and 



Average torque 


k I 0 E 0 C • T 

= T sm n 

J t =0 


(cot — 0 ) sin cot dt 


This leads by treatment similar to that employed for eq. 
262 to the expression 

average torpue = k E v I v cos 0. 


265 












176 


ELECTRICITY 


Readings therefore are proportional to true power. The pro¬ 
portionality constant, k, depends upon such factors as the di¬ 
mensions of the coils, the number of turns, etc. A wattmeter 
therefore after calibration reads true power for an a. c. circuit. 

The power factor of a circuit may be obtained by taking the 
ratio of true power, as given by a wattmeter, to the apparent 
power, as given by product of voltmeter and ammeter readings. 

235. Application of Ohm’s Law to A. C. Circuits. Because of 
the phase relations between e. m. f. and current some caution 
must be observed in applying Ohm’s Law to alternating current 
circuits. From eq. 252 we have 


Io 



This being true we may also write 
E v = ZI V . 

Since E v = Tg, and I v = A, 

V2 V2 


266 

any relation which exists be¬ 


tween maximum values must lead to a similar relation between 
virtual values. Thus E 0 2 = R 2 1 0 2 + L 2 co 2 1 0 2 or vectorially 
E 0 = R I 0 + Lcol 0 . Therefore E v 2 = R 2 I V 2 + L 2 co 2 I v 2 and 
vectorially E v = RI V + LcoI v . This might have been written 
directly from eq. 266. 

236. Addition of Electromotive Forces of the Same Frequency 
but out of Phase (Series Connection). The instantaneous value 

of the resultant voltage is 
ty the sum of the individual 

^ instantaneous values. These 


E, 




* 

A/WVW- 


instantaneous values are as 
we have seen the pro¬ 
jections of revolving radius 
vectors which have lengths 


Fig. 133 

equal to the maximum values and whose vector addition gives 
the resultant maximum voltage. The vector addition of the 
virtual or voltmeter values will give the virtual or voltmeter 
value of the resultant. Thus, vectorially, (E v ) r = (E v )i + (E v ) 2 



and, algebraically, 267 

(E v ) r 2 = (E v ) i 2 + (E v ) 2 2 -}- 2 (E v ) i (E v ) 2 cos0 

268 

where 0 is the phase difference between 
the two e.m. f’s. It will be noted that 
these virtual e. m. f s. are independent 
of time, i. e., thev are not the projec¬ 
tions of revolving vectors. 

The addition of electromotive forces of 









ALTERNATING CURRENT THEORY 


17 1 


different frequencies has important application in the radio tel- 
phone and will be discussed in that connection. 

237. Addition of Currents, (Parallel Connection). In this 

case the resultant current is 
the vector sum of the separate 
currents. Thus, vectorially, 
(Iv)r = (Iv)i + (Iv)a, 269 
and, algebraically, 

(Iv)r 2 = (Iv 2 )l+(Iv)2 

+ 2(Iv) x (Iv) 2 cos 0 270 

where 0 is the phase differ¬ 
ence of the currents. 0 = 
61 — 62 in which, as we have 
seen, we have only inductance 
and resistance. 


3i 

/ \ 




/V 


* 


Fig. 135 


0 i = tan 


■i iLi<o 


and ’02 = tan 


-i L2C0 . 


We should note carefully that the 


Ri-™ r 2 

resultant virtual current or voltage is not the arithmetrical sum 
of the separate currents or voltages unless they are in phase. 


238. Impedances in Series. 




Employing eq. 266 in eq. 267 
and cancelling the common 
factor, I v , we get, vectorially, 
Z r = Zi + Z 2 271 

Likewise, from 268 we get, 
algebracially, 272 


Fig. 136 


Since Z = 


Eo 

Io 


and since E 0 is 


made up of components (Lcol 0 +I 0 /cco)and R, (see Fig. 129). it 
followsjthat the components of Z r which may be due to inductance 

capacity or resistance may take the form Z = Leo or Z ^ or 

Z = R. 

A simplification is effected if we consider all series inductances 
as a single inductance, and similarly, all series capacities and 
resistances. A circuit will then have but three impedances and 
we may write, vectorially , 


= ( i + 


Lm) + R 


273 


inwhichC = 1/Ci+1/C2+ 
and R = Ri -f- R 2 -f- • • 


L = Li+ L 2 + 









178 


ELECTRICITY 


Equation 272 applied first to the inductance and capacity 
impedance gives numerically Z/ = ± (Zi — Z 2 ) since 6 is 180°, or 

Z' r = ± (Leo— J-). This is a numerical result. The resultant 
Coo 

mpedance is here stated to be the difference between the in¬ 
ductance and capacity impedances, each being considered as 
positive numbers. The difference is, by convention, considered 

positive if J- is the greater. Therefore the expression is usually 
Coo 


written Z' r 



— Leo). The matter of sign is purely arbitrary. 


Combining Z/with the series resistance impedance, Z 3 , we get, alge¬ 
braically, 


Z? = l'] + Z, 2 or Z? = (-L- La,) 2 + R 2 
Hence we have the general expression for impedance 

Zr = MH- RS 


This is identical’with 
eq. 253. 

239. Impedances 
in parallel. Since 
the resultant virtual 
current for two par¬ 
allel branches is (I v )? 
= (It): + (I»): + 
2(I v ) a (I v ), cos 0 where 
6 is the phase differ¬ 
ence between the cur¬ 
rents we get by em¬ 
ploying the relation 
E v = Z I v , 



1_ _ J_ j_1 _j_ 2 cos 6 

Z, f 2 z x 2 ' Z a 2 ‘ 7j\ 7j 


which may be written in the form 


Z 2 


Z, 2 Z 2 2 _ 

Z x 2 + Z 2 2 + 2 Z x Z 2 cos 6 


274 







179 


ALTERNATING CURRENT THEORY 


0 — 0i — 0a where 0i and 0 2 are the phase shifts of each current 

behind the applied e. m. f. 

In case there are several impedances in parallel the equivalent 
impedance may be found by combining the resultant of two with 
the third in accordance with eq. 274 and so on. 

240. Inductance and Capacity in Parallel, with Resistance 
Negligible. Employing eq. 274 and 
noting that 0=0, — 0 2 , and that 

0 ' = ten ' 1( - ^ “ I- and 01 = tan ' I( C^R } = + 

and that hence 0 = — x, we have, alge- 
braically, Z^- ZlZ l 

- Ln 


% 





• \y> - 

Pig. 138 


Since Z t = Leo and Z 2 = — we may 
write this 

La J- 

Z r = j—T" or Z r = 


Leo 


275 


Ca> Lc ° 


1 — LCco 2 


A case of great practical importance 

arises when ^-Leo = 0 in which 

Cco 


case LCeo 2 = 1 and Z r = 00 . The resultant current across ab, 
Fig. 138, and therefore everywhere in the main line is zero. The 

E E 

current through branch (1) is Ii = ^- = f-^, that through branch 

Li\ Leo 


P 1 

(2) is I 2 = — v = E v Ceo. Butlr|-I 2 = 0. Therefore I x =—1 2 ,and^- = 
Z 2 Leo 


Cco. The impedance around the loop, taking the inductance and 
capacity as in series, is therefore zero, see eq. 253. The frequency 
of the circuit may be obtained from the expression LCeo 2 =1 by 


writing 2xn for co, which gives n 


1 

2xVLCT' 


241. Oscillating Circuits. The loop composed of branches 
(1) and (2) Fig. 138, is a typical oscillating circuit. Since the 
impedance is zero, current once started flows and continues to 
flow in opposite directions in the two branches or around the 
loop, alternately charging and discharging the condenser. The 
energy of the circuit is at any instant partly that of the charged 
condenser expressed by JQ 2 /C, and partly that of the electromagnetic 
field, expressed by J LI 2 . There is no supply or expenditure of 












180 


ELECTRICITY 


energy, therefore the sum of these two forms must remain constant 
or 

iQ 2 /C+ i LI 2 = Const 276 

Placing I = ^ and differentiating with respect to time we have 


29 _Lq 

dt 2 LC y ' 


In this case^-^y 


47r 2 

co 2 = TpT, therefore we have 


47T 2 


277 


<PQ 

dt 2 T 2 

This is strikingly similar to the expression which characterizes 
simple harmonic motion for a mechanical system, viz., 
d 2 S _ _ 47T 2 

dt 2 T 2 S 

We may say at once therefore that the solution of 277 is 

Q = Qo sin cot 278 

Thus for an oscillating circuit the condenser charge oscillates 
between positive and negative maxima according to a sine law, 

278, and the current, which is oscillates or alternates be¬ 


tween positive and negative maxima according to the law 
I *=QoW cos cot. 

242. Coupled Circuits; The Transformer. When the chang¬ 
ing magnetic flux from one circuit sweeps across a neighboring 


circuit an e. m. f. is induced in that circuit equal to 


Nsd<£s 

KMt 



(paragraph 202). If the secondary circuit is closed, 
current flows under these conditions and energy appears in this 
circuit at the expense of that of the primary circuit. 
The equation which expresses the energy relations in the primary cir¬ 
cuit must be modified to take account of the effect of the secondary. 
Since, (Lenz’s law), the secondary current flows in a direction op¬ 
posite to that of the primary, its effect on the primary is to reduce the 
net magnetic flux cutting the primary coils and therefore*to 
reduce the effect of self-induction. We may view the effect as a 



partial neutralization of 
the back e. m. f. of self- 
induction. With the 
secondary circuit open 
only the inductance of 
the primary opposes the 
current. If the coil has 
an iron core enough 
current will flow to pro¬ 
duce the degree of mag- 














ALTERNATING CURRENT THEORY 


181 


netization necessary to make the opposing e. m. f. of self-induction 
very nearly equal to the applied e. m. f. The energy consumption 
of the circuit will be merely that required to produce the reversals 
of magnetization and to supply the R I 2 heat of the circuit. In 
case the secondary is closed energy appears in it at the rate E S I S or 


dl 


Mis and this must be absorbed from the primary. The 
energy equation for the primary circuit is given by 


E,!,—(L,«( I , + ) M f)l, + R,I ! 

That of the secondary is 

0= ( L! ^"( ’ +) M !r) l2+RsI ' 

The term, ( M -4jj-(,isthee. m. f. in the primary due to the mag¬ 
netic flux from the secondary cutting back across it. ' 

The corresponding e. m. f. equations are 

E 0 sin cot = Lx + M 4 r R 1 I 1 

at dt 


279 


0 = L 2 


dh 

dt 


+ M + R 2 I 2 


280 


The solution of these equations following the methods employed 
in paragraph 226 is tedious and need not concern us here, [see 
Starling, “Electricity and Magnetism”!. We will merely state 
as the result of such analysis, 

Ii = (Ii)o sin (cot— Oi) 281 

I 2 = ( 12)0 sin (cot— did-6) 282 

( 0 ) is the phase angle between the primary and secondary currents; 
(Oi), the phase angle between the primary e. m. f. and the primary, 
current, is given by the relation 

M 2 co 2 L 7 


di = tan 




L: 


LW +R? 


1 ) 


Also 

(Ii)o= 


Ri + 


E 0 


MVR, 
L*co 2 + R 2 


V (l, 


M^L, 


C + ) Ri + MWRj ■ ) 

\ / T, 2 (,$ 4-P 2 / 


283 


sin (cot— di) 
284 


Ly+R, 2 v / LjCo 2 + RJ- 

The effect of the secondary circuit on the impedance of the prima¬ 
ry and the angle of lag between primary e. m. f. and current is 
the same as though the inductance had been diminished by the 









182 


ELECTRICITY 


amount 


amount 


M 2 co 2 L 2 
L 2 co 2 +L 2 
M 2 co 2 R 2 

l 2 v+r 2 


and the resistance had been increased by the 


The e.m. f of the secondary circuit, which is M or ^ 

where is the rate of change of that part of the primary 

flux which cuts the secondary, is by Lenz‘s law, in such a direction 
as to oppose that change. If there were no self inductance or 
capacity in the secondary, the current would always be 180° 
behind the primary current in phase. The self-inductance and 
capacity of the secondary introduces an additional phase angle, 
a, so that we have 

I 2 = (Ia) 0 sin (cot —0i — 7r+a) 285 

where (—is the angle, 0, of eq. 282 
/I /C 2 C0—L 2 C0\ 

v i; ) 


a = tan - 1 


Ra / 286 

The analysis also shows that the relative value of maximum 
primary and secondary currents is given by the relation, 

(Do- — — - (10. 


V 


287 


LaC0* + R* 


ELECTROMAGNETIC RADIATION 


243. Electric Displacement. The theory of electromagnetic 

phenomena as elaborated by Clerk 
Maxwell is in close accord with 
- the accepted ideas of today. 
•k 4 4 ^ According to this theory a line 
^ of force (Faraday line) indicates 
an electric displacement in the 
dielectric surrounding a charged 
body. In the case of a charged 
1 condenser, for example, the elec- 

1 Fig. 140 11 -trostatic force exerted between the 

plates on application of a potential produces a displacement of 
the electricity of the dielectric throughout the region between the 
plates somewhat as pictured in Figure 140, I and II. The former 
indicates the condition of the dielectric when the condenser is 
uncharged, the latter when charged. A line of force extending 
across the dielectric from the positive to the negative charge 
indicates according to Maxwell the direction of the electric dis¬ 
placement or strain. 












ELECTROMAGNETIC RADIATION 


183 


244. Displacement Current or Movement of Faraday Lines. 

„ * » The condition above de- 

' \ scribed may be static as in 

' • the case of the condenser, 

1 ‘ in which case the electric 

field between the plates 
or about the charged body, 
may be observed by ex¬ 
perimental methods and the 
direction of the lines of 
force mapped out, or it 
may be kinetic as in the 
case of current flowing 
along a conductor. The 
latter may be illustrated 
by the circuit, Figure 
141, in which the cell main¬ 
tains a difference of potential 
across its terminals. On 
open circuit, assuming the 
connecting wire removed, 
the lines of force 
or displacement ex¬ 
tend from positive to nega¬ 
tive as indicated, each line 
originating at a positive 
charge and ending on the 
corresponding negative 
charge at the other terminal. 
When the cell is placed in 
a conducting circuit the 
ends of the force lines move 
with the positive and negative charges and collapse into the 
conductor as the charges neutralize. The magnetic field which 
accompanies any flow of current is thought of as associated with 
the moving electric force lines and the energy which the magnetic 
field represents as the kinetic energy of the moving force lines. 
As these lines collapse into the conductor their energy is absorbed 
into the conductor as heat. 

245. Electric Oscillations and Radiation. Figure 142 repre¬ 
sents a so-called Hertzian 
radiator, composed of spark- 
gap, condenser plates and 
connecting wires. The con¬ 
denser plates are connectep 
to the secondary of an 
induction coil. The dotted 
lines represent the electric 



Fig. 141 


• .1 



Fig. 142 


















184 


ELECTRICITY 


field or the lines of force about the gap when the plates are charged. 
If the voltage is raised to the sparking potential a discharge will 
occur across the gap. In the case of long wires connecting con¬ 
denser plates and gap or wires of high resistance the flow of 
current will be retarded and the lines will collapse upon the gap 
as the charges flow together. With low resistance wires the ve¬ 
locity of electron flow is greater and the ends of the force lines 
may reach the center of the gap before the more remote portions 
have had time to move all the way in. In this case the inertia 
of the moving lines carries these ends past each other and the 
condenser charges in the opposite direction. When the inertia 
is spent and the energy reconverted into the electrostatic energy 
of the charged condenser, the charges surge back again across the 
gap. This continues while the energy is slowly dissipated as 
heat in the conducting wires and gap. The frequency of these 
oscillations is determined by the constants of the circuit and is 


given by the expression, n 


1 

2*VLC 


(eq. 256). 


When the ends of a force line have crossed, a loop will be formed 
of the outlying portion as shown, Figure 143, (I). Near the 
point of intersection the lines will be moving parallel to their 
length and will possess no inertia, for a magnetic field is associ¬ 
ated only with Faraday lines moving at right angles to their 
length. These portions of the lines therefore will not be carried 
past one another but will coalesce. The motion of this part of 
the loop toward the gap has ceased, other loops are forming 
outside and around it with a crowding of lines between it and the 
gap. The loop is thus repelled from the gap. We now have the 
conditions of Figure 143, (II), and (III). As the other lines 
move in they snap off in loops around the first as shown in these 
figures. Figure 143 (IV) shows the force lines which are es¬ 
tablished as the condenser becomes charged in the opposite 
direction, together with the first set of loops now entirely free 
from the gap and, under the repulsion of the lines behind, travel¬ 
ing away along the axis. We thus see a series of Faraday lines 



Fig. 143 
















ELECTROMAGNETIC RADIATION 


185 


or electric displacements traveling outward along the axis or 
perpendicular of the gap, the displacements being at right angles 
to the axis or direction of travel and alternating in sign with the 
frequency of the oscillating current. Theory and experiment 
indicate that these lines travel away with the velocity of light. 



A spark gap is not an 
essential to electromagnetic 
radiation. Any circuit in 
which sufficiently high-frequen¬ 
cy oscillations are maintained 
will radiate in the same way, 
Figure 144. 

Fig. 144 

246. The Detection of Electromagnetic Waves. As has 

been stated a magnetic field is associated with moving Faraday 
lines, its direction being at right angles to these lines. Any 
relative motion between a conductor and a magnetic field results 
in a current flow in the conductor. Therefore any conductor 
which forms a part of a closed circuit placed in the path of electro¬ 
magnetic waves and so oriented that there is a component of the 
magnetic field at right angles to the conductor will be subjected 
to a high-frequency electromotive force. The current in such a 
receiving circuit will be a maximum if the constants of the circuit 
are such that its natural frequency coincides with that of the 

waves. The frequency is given bv the expression n = ——^ 7 =, 

^VLC 


and the tuning, as it is called, is effected by adjusting the capacity 
or the inductance or both. 

There are innumerable ways of arranging oscillating circuits 
both for sending out electromagnetic waves and for receiving 
them. Figure 145 (I) and (II) indicate possible arrangements. 



Fig. 145 






















186 


ELECTRICITY 


Oscillations are set up in the sending circuit, a, by the three- 
electrode vacuum tube whose action is explained below. This 
circuit is inductively coupled to circuit, b, where the oscillations 
are amplified, (also by the action of the vacuum tube as will be 
shown), and passed by inductive action to the antenna circuit, c. 
The frequency is determined by adjustment of the capacity, ci, 
in circuit, a. The antenna frequency is tuned to this by ad¬ 
justment of inductance, L 4 . If the receiving antenne, Figure 
145 (II) is in the path of the waves and is tuned to the same 
frequency, oscillations will be set up in it and transferred by 
inductive action to the tuned oscillating circuit of the receiving 
set. It will be necessary to discuss the action of the thermionic 
vacuum tube before the further mechanism of the reception of 
signals can be understood. 

247. The Three-Electrode Vacuum Tube. In the very 
early incandescent carbon lamps a third electrode not connected 
with the carbon filament was for some reason always left sealed 
into the bulb. Edison noticed that when the positive side of a 
battery was connected to the extra lead and the other side to 
either of the filament leads a current would flow whenever the 
filament was heated. The direction of this current was always 
toward the hot filament. How the current got across the vacuum 
from the extra lead to the filament was not explained. This 
“Edison effect” as it was called was regarded as a curious phe¬ 
nomenon but at the time received little attention or study. An 
explanation was found many years later by Richardson and 
others in the course of an investigation of the emission of electrons 
or negative particles of electricity from hot bodies. If ab, 

Figure 146, represents a hot filament 
and, c, a conducting plate, both 
enclosed in a vacuum bulb, it is 
found that, whenever the contact 
is closed, current will flow through 
the galvanometer. This is merely 
the “Edison effect.” The explanation 
is found in terms of the electron 
theory, viz., that with sufficient 
thermal agitation electrons escape 
through the surface of the metal and 
are attracted to any conducting body 
plate in the neighborhood which thereby 
becomes negatively charged with respect to the filament. When the 
plate is connected externally to the filament current flows. Current 
is carried across the gap by means of the escaping electrons. 

This thermonic current, as it is called, is found to increase with 
the filament temperature according to an exponential law which 










ELECTROMAGNETIC RADIATION 


187 


is of exactly the same form as that which holds for the liberation 
or vaporization of molecules through the surface of a liquid, thus 
suggesting that the electrons within the hot filament behave 
like the free molecules within a liquid. We know that above 
the liquid there is a vapor formed by the molecules which have 
escaped through the surface, and just so in this case there appears 
to exist about the hot filament a vapor of electrons which have 
escaped through the surface. When the electron vapor com¬ 
pletely fills the space between the filament and plate, the further 
emission of electrons from the filament is checked by the re¬ 
pulsion due to this so-called “space charge.” If the plate is 
maintained at a positive potential by means of a battery, the 
passage of electrons will be facilitated by the greater attraction 
of the plate and the thermonic current correspondingly increased. 
If the plate voltage is increased to such an extent that all the 
electrons are pulled over to it as fast as they are emitted the 
current is then a maximum for the particular temperature of the 
filament and is called the saturation current. The relation 



between the plate current and the plate voltage for two filament 
temperatures and also the relation between the plate current 
and the filament temperatures for two plate voltages is shown 
in Figure 147 I and II. 








188 


ELECTRICITY 


If a metallic mesh or grid 
is placed between the filament 
and plate, (Fig. 148), and 
« positively charged, the effect 

* of the space charge will be 

* more or less neutralized de- 

. pending upon the potential 

* of the grid. The electrons 
which fill the space in front 
of the filament will be at¬ 
tracted to the grid, those 
that strike the grid will be 
led away, while those that 
pass through the meshes will 
travel on toward the plate if 
the plate potential is main¬ 
tained higher than that of 
the grid. If the grid is nega¬ 
tively charged the flow of electrons may be entirely shut off. 
Figure 149 shows how the plate current is effected by the grid 



'F 



voltage. If the grid potential 
is zero the plate current has 
the value it would have for 
the two electrode tube. With 
negative grid potential the 
flow of electrons is retarded 
and the plate current is less, 
with positive potentials the 
space charge is neutralized 
and the flow of electrons to 
the plate is increased until 
some particular grid voltage 


allows the maximum current to flow for the given filament temper¬ 
ature. Curves showing these relations (Figures 147, 149) are 
called the characteristic curves for the tube. 


If the grid voltage is adjusted to some point, x, Figure 149 
such that we are operating on the steep portion of the curve 
then slight changes in grid voltage lead to large changes in plate 
current. The tube is then in condition to act as an amplifier. 
If we adjust the grid voltage so that we operate near some point, 
y, Figure 149, then increments of grid voltage produce greater 
effects than decrements and we are in condition to operate the 
tube as a detector. 










ELECTROMAGNETIC RADIATION 


189 


Figure 150 shows a typical vacuum tube circuit. The potential 

of the grid may be adjusted 
by the “C” battery until 
the desired region of the 
curve is reached. If an 
oscillating e. m. f. is applied 
to the grid circuit through 
the coupling, Li, periodic 
fluctuations in the grid po¬ 
tential are produced and 
corresponding fluctuations of the plate current result. It should 
be noted that the direction of the plate current is always the same, 
the action of the grid merely producing fluctuations in its value. 

248. The Vacuum Tube as a Detector. The action of the 
detecting circuit shown in Figure 145 (II) may now be under¬ 
stood. The antenna oscillations excite corresponding oscillations 
in the grid circuit which is tuned to resonance by adjustment of 
condenser, Ci. These oscillations produce variations of the 
grid potential. Fluctuations in the intensity of the plate current 
are thus set up. If we operate the tube in the region of, y 
Figure 149, which may or may not be at zero grid voltage, de¬ 
pending upon the shape of the characteristic for the particular 
bulb, the diminutions of grid voltage do not produce as great 
fluctuations of plate current as do the increments, Fig. 151. If now 
the oscillations are interrupted either at the sending or receiving 




station and the number of interruptions per second corresponds 
to some audible note, a sound will be heard in the telephone. 
This response is due to the fact that while the telephone cannot 
follow the radio-frequency oscillations it can respond to the 
group impulses which are of audio-frequency. If we operate at 
point, x, there will be no average increase or decrease in plate 
current between interruptions and the telephone will not respond. 
If a series of damped waves are sent out as from a spark dis¬ 
charging at audio frequency the plate current will vary as shown 
n Figure 152 and the telephone will respond exactly as before 












190 ELECTRICITY 



249. The Vacuum Tube as an Amplifier. If the tube is 
operated a£ a point near x, Fig. 149, the fluctuations of the 
plate current about the mean will be symetrical and very much 
greater for a given change in grid voltage than when operated 
at y. A given change in voltage applied to the grid will produce 
six or seven times the change in the plate current that one would 
get if the same voltage were applied to the plate. Figure 152 
shows an amplifying set added to the detecting set of the receiving 
circuit shown in .figure 145, (II). The vacuum tube of the de¬ 



tecting set is operated at point, y, of the characteristic, that of 
the amplifying set at point, x. The constants of the tuned 
circuit (a) in the detecting set must be chosen for radio-frequency, 
those of the tuned circuit (b) in the amplifying set for audio¬ 
frequency. An idea of the extent of amplification may be ob¬ 
tained from the fact that a change of grid current in the ampli¬ 
fying bulb of one microampere may lead to a change in the plate 
^current of 100 microamperes. If the grid voltage is two volts 
and the plate voltage sixty, the power amplification amounts to 
3,000 times. This power of course is drawn from the plate 
battery. 






























ELECTROMAGNETIC RADIATION 


191 


250. The Vacuum Tube as a Generator. If the plate circuit 
is coupled back to the tuned circuit (a), Figure 154, which is 
connected to the grid, oscillations in (a) will be amplified in the 
plate circuit and impressed back upon (a) through the coupling, 



thus building up and maintaining oscillations in (a). These 
oscillations may be communicated to a tuned antenna circuit 
through coupling with the plate circuit. 

251. Heterodyne Reception by Vacuum Tubes. If sustained 
oscillations are communicated to the receiving circuit (a), Figure 

155, simultaneously from 
both the antenna and 
a local oscillating circuit 
(b) of slightly different 
frequency, there will be 
produced in the tele¬ 
phone a beat note which 
may be made of audible 
frequency. The super¬ 
position of two such 
alternating currents is 
shown graphically in 
Figure 156. The result¬ 
ing current (instanta¬ 
neous value) is given as 
the sum of the two cur¬ 
rents thus, 




































192 ELECTRICITY 

I = Ii sin 27rnt + I 2 sin 27r(n + f)t 288 



It happens that the action of the telephone mechanism depends 
upon the square of the current. This is of great advantage as 
it enables us to operate the tube in the amplifying region. Any 
device which responds to the first power of a radio-frequency 
current requires distortion of the wave-form, such as is produced 
by a tube operating in the region near y, Fig. 149, in which 
case the increments of grid voltage have greater effect on the 
plate current than the decrements. With the telephone how¬ 
ever we have the advantage of detection and amplification at 
the same time. The presence of an audio-frequency component 
to which the telephone can respond is indicated by the mathe¬ 
matics of the case. Thus, squaring eq. 288 we have, 

I 2 = Ii 2 sin 2 27rnt + I 2 2 sin 2 27r(n + f)t 

+ 2 Ii I 2 sin27mt-sin 27r(n + f)t. 

Remembering that sin 2 a = i(l—cos 2 a) and that 
sin a sin (3 = ^ cos (a —cos ( a + /3), we have 

I 2 = ^ cos 47 rnt+ — cos 47 r (n + f)t 

— Iil 2 cos 27rft—IJ 2 cos 27r(2n + f)t. 290 

We have here zero frequency terms, terms of radio-frequencies, 
n and (n + f), a term which is the sum of these two and therefore 
also of radio-frequency, and a term which is their difference and 
therefore of frequency, f. f may be made of an audio-frequency 
to which the telephone can respond. If, for example, the received 
waves have a frequency 500,000 and the local circuit 501,000, 
the telephone will pick up the beat note of 1,000. The reception 
of radio signals by means of a beat note caused by the interference 
of locally generated oscillations is known as the heterodyne 
method. If the two frequencies are sent out by the generating 
station, heterodyning does not have to be resorted to at the 
receiving end and the local oscillating circuit at the receiving 
end is unnecessary. 






ELECTROMAGNETIC RADIATION 193 

252. Radio-Telephony; Voice Modulation. If an antenna 
to which is being communicated radio-frequency oscillations is 
also coupled to the plate circuit of a tube whose grid potential is 





Fig. 157 


Fig. 158 



varying, the intensity of the oscillations induced in the plate by 
the antenna will depend upon the effective resistance of the plate 
circuit. The plate or tube resistance depends upon the facility 
with which electrons get from filament to plate and this depends 
upon th3 grid potential. With negative grid potential the re¬ 
sistance is very high and the plate circuit absorbs very little 
energy from the antenna. With lower tube resistance which 
results from the grid potential becoming less strongly negative or 
even positive more of the radio frequency current is induced in 
the plate coil and thereby more energy absorbed from the antenna. 
If the grid potential follows the varying frequency of the voice 
in the telephone, Figure 157 and 158, the intensity of the radio 
frequency oscillations of the antenna will correspondingly follow 
the voice modulations. In other words the grid potential which 
follows all the fluctuations of the voice, Figure 158, controls the 
absorption of energy by the plate circuit from the antenna. 


















194 


ELECYRICITY 


The intensity of the an¬ 
tenna oscillations are 
therefore modulated in 
exact accordance with 
the voice in the tele¬ 
phone. In the receiving 
set, Figure 159, the modu¬ 
lated oscillations pro¬ 
duce corresponding mod¬ 
ulations of the grid po¬ 
tential and of the recti¬ 
fied plate current, the 
telephone finally repro¬ 
ducing the original voice 
modulations. The radio¬ 
frequency whose intensi¬ 
ty is thus varied is called 
the carrier frequency. 
There are numerous arrangements of circuits used in radio 
work, many of them very complicated but all based upon the 
same fundamental principles. 



Fig. 159 













INDEX 


References are to pages. 


Aberration, chromatic, 97, 98 
spherical, 86, 97 
Absolute zero, 44 
Absorbing power, 72 
Acceleration, 2 

Achromatic combination, 98 
Action and reaction 8 
Adiabatic changes, 58 
equation of, 58 
Alternating current, 165 
Ammeter, 139 
Ampere, 129, 135 
Ampere—turn, 157 
Amplifier, vacuum tube as, 190 
Angstrom unit, 92 
Angular motion, 23 
Armature, ring, 143 
drum, 144 
Avogadro’s law, 98 
Axis, optic, 109 

of a crystal, no 

Black-body, 72,73 
Boiling point, 54 
Bolometer, 74 
Boyle’s law, 43, 41 
British thermal unit, 51 
Brownian movements, 49 

Calorie, 51 
Capacity, 115, 117 
of sphere, 117 
of plate condenser, 118 
of spherical condenser, 118 
of cylindrical condenser, 119 
Carnot cycle, 60 
Carnot engine, 60 
Carnot’s theorem, 62 
Carrier frequency 194 
Caustic curve, 84 
Cell, electrolytic, 161 
Daniell, 164 
polarized, 162 
storage, 165 
voltaic, 162 
Centimeter, 8 
Center of gravity, 28 
Center of mass 25, 26 
Charles’ law, 43 
Color, 85, 100 


Collimator, 89 
Condenser, 116 
energy of, 120 
Conduction, 71 
Convection, 71 
Coulomb’s law, 112, 121 
Couple, 27 

Coupled circuits, 180 
Critical angle, 85 
point, 55 

Current, average AC, 172 
virtual AC, 172 

Detection of electric waves, 185 
Detector, vacuum tube as, 189 
Deviation, minimum, 88 
without dispersion, 91 
Dielectric constant, 116 
Diffraction, 104 
Dimensions, 10 
Dimensional equations, 10 
Dispersion, 87, 90, 91 
Dispersive power 90 
Displacement, 1 
Displacement current, 183 
Displacement, electric, 182 
Dynamo, a.c., 141 
d.c., 142 
Dyne, 8 

Edison effect, 186 
Efficiency of a heat engine, 61, 70 
Elasticity, 37 
Elastic limit, 38 
Electric current, 127, 129 
field, 113 

Electro—chemical equivalent, 160 
Electrolysis, 160 
Electromotive force, 135 
E.M.F. vertual A.C., 172 
Electromagnetic theory, 84 
radiation, 182, 183 
waves, 185 
Electrons, 186 
Emission of electrons 

from hot bodies, 186 
Emissivity, 72 
Energy, 20, 21, 31 
Entropy, 65 , 66 
Erg, 18 


INDEX 


Ether, 84 
Expansion, 42 

Farad, 117 
Faraday, 140 

lines, movement of 183, 184 
Faraday’s laws, 160 
Field due to current, 128 
at center of loop, 129 
on axis of coil, 129 
near straight wire, 130 
within solenoid, 131 
Flourescence, 93 
Focal length, 81 
Focus, principal, 81 
conjugate, 81 
depth of, 84 
Foot-pound, 18 
Force, 7 

Force, on wire carrying current, 126 
Force, line of 121 
Fraunhofer lines, 93 
Frequency of AC circuit, 179, 185 
Fresnel zones, 104 
Fringes interference, 105 
Fundamental interval, 41 
Fusion, 54 

heat of, 54 

Galvanometer, ballistic, 149 
D’Arsonval, 132 
tangent, 132 
Gas laws, 43 
Gases, properties of, 56 
Gauss’s theorem 113, 122 
Generator, vacunm tube as, 191 
Gi am, 8 

Gramme ring, 143 
Grating, diffraction, 106 
Grid, 188 

voltage, 188 
Gyroscopic motion, 35 

Heat capacity, 51 
Heat, units of, 51 
specific, 52 

Heterodyne reception, 191 
method 192 
Hooke’s law, 37 
Horse-power, 18 
Huyghen’s principle, 85 
Hydrogen thermometer, 68 
Hysteresis, 154, 155 

Image in plane mirror, 86 
virtual, 97 

Impedances in series, 177 
in parallel, 178 


Impulse of a force, 7 
Impulse of a torque, 30 
Impedance, 168 
Induced e.m.f., 140 
Inductance, 147 
Induction, electric, 113 
magnetic, 121, 153 
self, 147 
mutual 148 

Interference of light 85, 109 
Isothermal, 55 
changes, 57 

Joule, 18 
Joule’s law, 134 

Kilowatt-hour, 18 
Kinetic theory of gases, 46 
Kirchhoff’s laws, 136 
radiation law, 72 

Lens, converging, 94, 95 
diverging, 94, 96 
path of ray through, 87 
Lenses, 94, 97, 98 
L'enz’s law, 140 
Light as wave motion, 84 
Liquifaction of air, 51 
Logarithmic decrement, 151 
Luminescence, 93 

Magnetic circuit 157 
field, 121 

induction, 121, 153 
leakage, 125 
moment 126 
Magnetism, 121 
Magnetized iron ring, 125 
iron bar, 125 

Magnetomotive force, 157 
Magnification, 99 

Mechanical equivalent of heat, 52 
Micron, 92 
Microscope, 100 
Mirror, concave spherical 81 
convex spherical, 82 
parabolic, 83 
Modulation, voice, 193 
Modulus of elasticity, 37 
Molecular attractions, proof of, 50 
Moment of force, 24 
of inertia, 29 
Momentum, angular, 30 
Momentum, linear, 7 
Motion, gyroscopic, 35 
Newton’s laws, of, 6 
of a projectile, 4 
of a rolling body, 32 


INDEX 


simple harmonic, 12 
uniform circular, 5 
uniformly accelerated, 3 
Motor, d.c., 144 

efficiency of, 145 
induction, 146 

Newton’s .law of cooling, 71 
Nicol prism, no 
Nitrogen thermometer, 68 

Ohm, the 135 
Ohm’s law, 134,137,148 
in a.c. circuit, 176 
Optic, axis, 109 
Optical instruments, 99 
Oscillating circuits, 179, 185 
Oscillations, electric, 183 

Pendulum, Kater’s 34 
physical, 34 
torsion, 39 
Perfect gas, 43 
Permanent set, 38 
Permeability, 126 
Phosphorescence, 93 
Planck’s law, 75 
Plane of polarization, 109, m 
Plate current 188 
Plate voltage, 188 
Polariscope, in 
Polarized light, 107, 109 
Power, 18 

factor, 173, 174 
in electric circuit, 140, 173 
Porous plug experiment, 50 
Potential, electric, 114, 115 
magnetic, 127 
Potentiometer, 138 
Pressure, expression for, 47 
Prism, 87 
nicol, no 
Process, cyclic, 59 
reversible, 59 
irreversible, 59 
Pyrometer, optical, 75, 76 
—Fery, 74 

Quantity of electricity, 132 

Radiation, 71 
Radio-telephony, 193 
Range of a projectile, 4 
Reactance, 168 
Rectilinear propagation, 85 
Reflection at plane surface, 86 
Reflection at spherical surface, 87 


Refraction, 85, 86, 87 
index, of, 85 

Refrigeration, mechanical, 51 
Reluctance, 157 
Resistance, electrical, 134 

temperature coefficient of, 139 
Resistances in parallel, 139 
Richardson, 186 
Rigid body, motion of, 27 
Rigid bodies, equation of motion, 32 
Root mean square current, 173 
Root mean square e.m.f., 173 

Sagitta theorem, 81 
Saturation current, 187 
Scalar quantities, 1 
Shaft, twisted, 38 
Shear, 37 

Signs, rule of, J82, 96 
Simple harmomic motion, 12 
Snell’s law, 85 
Solenoid, 131 
Sound, velocity of, 11 
Space charge, 187 
Spectra, 91 
Spectrometer, 89 
Spectroscope, direct vision, 91 
Spherical aberration, 84, 97 
Spring, 16, 19 
State, equation of, 44 
change of, 53 
Stefan’s law, 73 
Strain, 37 
Stress, 37 

Submarine cable, capacity of, 119 
Susceptibility, 126 

Telescope, astronomical, 100 
Galileo’s, 102 
Temperature, absolute, 44 
centigrade, 41 
Fahrenheit, 41 
perfect gas, 44 
scale, 41 

thermodynamic, 59 
Tensile strength, 38 
Thermal conductivity, 71 
Thermodynamic equations, 68 
Thermodynamics, first law of, 52 
Thermodynamics, second law of, 62 
Thermo-electricity, 165 
Thermojunction, 165 
Thermometer, 41, 48 

Torque, 30 

Torque on a magnet, 126 
Tourmaline, 108 
Transfer of heat, 71 
Transformer 185 


INDEX 


Uniform temperature enclosure, 73 
Units, 8, 9 
Units, absolute, 9 
Units, c. g. s., 8 

Vacuum Tube, 3-electrode, 186 
Van der Waal’s equation, 49 
Vapor pressure, 53 
Vapor, saturated, 53 
Vaporization, 53 
heat of, 53 
Vector, 1 

diagram for a. c. circuit 171 
Vectors, in a.c. circuit, 169 
angular, 24 
Velocity, 2 
Volt, the, 135 
Voltage, virtual 172 
Voltmeter, 139 


Watt, 18 
Wattmeter, 175 
Wallaston, 93 
Wave-length units, 92 
Weight and mass, 9 
Wheatstone bridge, 138 
Wien’s displacement law, 74 
Wien’s distribution law, 75 
Work, 17, 18, 30, 45 
Work on the p v diagram, 45 


Yield point, 38 
Young, Thomas, 85 
Young’s experiment, 85 
Young’s modulus, 37 


Zone plate, 102 






















































































































































































































































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